Integrals

Since d(cos 2x)/dx=−2sin 2x, divide by −2.

Since d(sin 3x)/dx=3cos 3x, divide by 3.

The derivative of e^(2x) is 2e^(2x).

Divide by 2 to cancel the extra factor.

The derivative of (ax+b)³ is 3a(ax+b)².

Divide by 3a (when a≠0).

Integrate each term separately.

∫sin 2x dx=−(1/2)cos 2x and ∫4e^(3x)dx=(4/3)e^(3x).

Let u=4x⁴+1. Then du=16x³dx, so x³dx=du/16.

Substitute and integrate: (1/16)∫eᵘdu.

Rewrite x²/(1−x²)=−1+1/(1−x²).

Use 1/(1−x²)=1/2·[1/(1+x)+1/(1−x)].

Integrate each simple fraction.

Integrate each term using the power rule.

Expand the polynomial factor: x²(x+2)=x³+2x².

Use repeated integration by parts, or check by differentiating eˣ(x³−x²+2x−2).

Integrate 1/x as ln|x|.

Integrate −x as −x²/2.

Divide each numerator term by x²: x−5+4x⁻².

Integrate term by term.

Divide each term by x: x²+1+4/x.

Integrate the polynomial terms and the logarithmic term.

Factor by grouping: x³−x²+x−1=(x−1)(x²+1).

Cancel x−1 (on the domain x≠1), leaving x²+1.

Integrate the remaining polynomial.

Expand: (1−x)x=x−x².

Integrate each power.

Expand: x⁵+2x³+3x².

Integrate term by term.

Integrate the first two terms directly.

For xeˣ, use integration by parts with u=x and dv=eˣdx, giving ∫xeˣdx=eˣ(x−1).

Integrate each term separately.

Recognise that d(tan x+sec x)/dx=sec²x+sec x tan x.

Since 1/cosec²x=sin²x, the integrand is sec²x·sin²x=tan²x.

Use tan²x=sec²x−1.

Split the fraction: 2sec²x−3tan²x.

Use tan²x=sec²x−1, so this becomes 3−sec²x.

Set t=x^(1/6), so x=t⁶ and dx=6t⁵dt.

Then (1/(√x+∛x))dx=6t³/(t+1)dt=6(t²−t+1−1/(t+1))dt.

Integrate these four terms and substitute t=x^(1/6).

Rewrite f′(x)=1/x−4/x⁴.

Integrate: f(x)=ln x+4/(3x³)+C (the condition x=2 gives x>0).

Use f(2)=0: C=−ln 2−1/6.

Let u=1+x², so du=2x dx.

The integral becomes ∫du/u.

Let u=log x, so du=dx/x.

The integral becomes ∫u²du.

Let u=log x, so du=dx/x.

Then the integral is ∫du/u.

Let u=x+2. Then x=u−2 and du=dx.

Rewrite the integral as ∫(u−2)u^(1/2)du.

Apply the power rule to u^(3/2)−2u^(1/2).

Let u=cos x. Then du=−sin x dx.

The integral becomes −∫sin u du=cos u+C.

Let u=sin(ax+b), so du=a cos(ax+b)dx.

Then the integral is (1/a)∫u du, when a≠0.

Let u=ax+b, so du=a dx.

For a≠0, the integral becomes (1/a)∫u^(1/2)du.

Let u=1+2x², so du=4x dx.

Rewrite as (1/4)∫u^(1/2)du and apply the power rule.

Let u=x²+x+1, so du=(2x+1)dx and (4x+2)dx=2du.

The integral becomes 2∫u^(1/2)du.

Set t=√x, so x=t² and dx=2t dt.

Then the integral becomes 2∫dt/(t−1).

Set u=x+4, so x=u−4 and dx=du.

Integrate (u−4)/√u=u^(1/2)−4u^(−1/2).

Let u=x³−1, so du=3x²dx.

The integral is (1/3)∫u^(1/3)du.

Let u=2+3x³, so du=9x²dx.

The integral becomes (1/9)∫u^(−3)du.

Let u=log x, so du=dx/x.

The integral becomes ∫u^(−m)du.

Let u=9−4x², so du=−8x dx.

The integral becomes −(1/8)∫du/u.

The derivative of 2x+3 is 2.

Divide by 2 to reverse the chain rule.

Rewrite as xe^(−x²). Let u=−x², so du=−2x dx.

Let u=tan⁻¹x, so du=dx/(1+x²).

The integral becomes ∫eᵘdu.

Set t=e^(2x), so dx=dt/(2t).

The integral becomes (1/2)∫(t−1)/[t(t+1)]dt=(1/2)∫[−1/t+2/(t+1)]dt.

Differentiate ln(e^(2x)+e^(−2x)); the chain rule gives twice the numerator divided by the denominator.

Therefore the required antiderivative is half that logarithm.

Use tan²u=sec²u−1.

Let u=2x−3, so dx=du/2.

Let u=7−4x, so du=−4dx.

Use ∫sec²u du=tan u.

Let u=sin⁻¹x. Then du=dx/√(1−x²).

The integral becomes ∫u du.

The denominator’s derivative is −6sin x+4cos x=2(2cos x−3sin x).

So the integrand is one half the derivative of the denominator divided by the denominator.

Set t=tan x, so dt=sec²x dx=dx/cos²x.

The integral becomes ∫dt/(1−t)².

Let t=√x, so dx=2t dt.

The integral becomes 2∫cos t dt.

Set u=sin 2x, so du=2cos 2x dx.

The integral is (1/2)∫u^(1/2)du.

Set u=1+sin x, so du=cos x dx.

Integrate u^(−1/2).

Let u=log(sin x). Then du=cot x dx.

The integral becomes ∫u du.

The derivative of 1+cos x is −sin x.

Use the logarithmic derivative rule.

Let u=1+cos x, so du=−sin x dx.

The integral becomes −∫u^(−2)du.

Rewrite 1/(1+cot x)=sin x/(sin x+cos x).

Write sin x as one half of (sin x+cos x)+(sin x−cos x).

Integrate the constant part and use that sin x−cos x is the negative derivative of sin x+cos x.

Rewrite as cos x/(cos x−sin x).

Write cos x as one half of (cos x−sin x)+(cos x+sin x).

The second sum is the negative derivative of cos x−sin x.

Set t=tan x, so dt=dx/cos²x. Also sin x cos x=tan x cos²x.

The factors simplify the integral to ∫t^(−1/2)dt.

Let u=1+log x, so du=dx/x.

The integral becomes ∫u²du.

Let u=x+log x. Then du=(1+1/x)dx=(x+1)dx/x.

The integral becomes ∫u²du.

Let u=tan⁻¹x⁴. Then du=4x³dx/(1+x⁸).

The integral becomes (1/4)∫sin u du.

Differentiate the denominator x¹⁰+10ˣ. Its derivative is exactly 10x⁹+10ˣlogₑ10.

Apply ∫f′(x)/f(x)dx=log|f(x)|+C.

Use 1/(sin²x cos²x)=1/sin²x+1/cos²x, since sin²x+cos²x=1.

Integrate cosec²x and sec²x.

Use sin²u=(1−cos 2u)/2.

With u=2x+5, integrate 1/2−(1/2)cos(4x+10).

Use sin A cos B=[sin(A+B)+sin(A−B)]/2.

The product becomes [sin 7x−sin x]/2.

Integrate both sine terms.

First use cos 2x cos 6x=[cos 8x+cos 4x]/2.

Then multiply by cos 4x and use cos A cos B=[cos(A+B)+cos(A−B)]/2.

The product becomes [1+cos 8x+cos 4x+cos 12x]/4.

Write sin³u=sin u(1−cos²u).

Let t=cos(2x+1), so dt=−2sin(2x+1)dx.

Integrate −(1/2)(1−t²)dt.

Write sin³x cos³x=sin x(1−cos²x)cos³x.

Let t=cos x, so dt=−sin xdx.

Integrate −(1−t²)t³.

Use sin x sin 3x=(cos 2x−cos 4x)/2.

Then use sin 2x cos 2x=sin 4x/2 and sin 2x cos 4x=(sin 6x−sin 2x)/2.

The product simplifies to (sin 4x−sin 6x+sin 2x)/4.

Use sin A sin B=[cos(A−B)−cos(A+B)]/2.

The integrand becomes (cos 4x−cos 12x)/2.

Use the half-angle identity (1−cos x)/(1+cos x)=tan²(x/2).

Use tan²u=sec²u−1 and integrate with u=x/2.

Rewrite cos x/(1+cos x)=1−1/(1+cos x).

Since 1+cos x=2cos²(x/2), an antiderivative of 1/(1+cos x) is tan(x/2).

Use sin²x=(1−cos 2x)/2 and square.

After simplifying, sin⁴x=(3−4cos 2x+cos 4x)/8.

Use cos⁴u=(3+4cos 2u+cos 4u)/8 with u=2x.

So cos⁴(2x)=3/8+(1/2)cos 4x+(1/8)cos 8x.

Factor sin²x=(1−cos x)(1+cos x).

Cancel 1+cos x where the original expression is defined, leaving 1−cos x.

Use cos 2x=2cos²x−1 and cos 2α=2cos²α−1.

Factor: cos 2x−cos 2α=2(cos x−cos α)(cos x+cos α).

Cancel the common factor on intervals where the original quotient is defined.

Use 1+sin 2x=(sin x+cos x)².

The numerator is the derivative of sin x+cos x.

Set u=sin x+cos x; then integrate du/u².

Let u=tan(2x), so du=2sec²(2x)dx.

Then the integral is (1/2)∫u³du.

Use tan²x=sec²x−1 to write tan⁴x=(sec²x−1)².

For ∫sec⁴x dx, write sec⁴x=(1+tan²x)sec²x and use u=tan x.

Thus ∫sec⁴x dx=tan x+tan³x/3; subtract 2∫sec²x dx and add ∫1dx.

Use sin²x+cos²x=1.

Integrate sin³x=sin x(1−cos²x) with u=cos x.

Integrate cos³x=cos x(1−sin²x) with u=sin x.

Split the fraction: 1+2tan²x.

Use tan²x=sec²x−1, giving 2sec²x−1.

Let t=tan x. Then dx=dt/(1+t²), while sin³x cos x=t³/(1+t²)².

The integral becomes ∫(1+t²)/t³ dt=∫(t⁻³+t⁻¹)dt.

Set t=tan x. The integral becomes ∫dt/[(t+1)²(1+t²)].

Use partial fractions: 1/[(t+1)²(1+t²)]=1/[2(t+1)]+1/[2(t+1)²]−t/[2(1+t²)].

Integrate each term and substitute t=tan x.

On this interval, cos x=sin(π/2−x), and π/2−x lies in the principal range [−π/2,π/2].

Therefore sin⁻¹(cos x)=π/2−x.

Integrate this linear expression.

Use tan(x−a)−tan(x−b)=sin(b−a)/[cos(x−a)cos(x−b)].

Rewrite the integrand as [tan(x−a)−tan(x−b)]/sin(b−a).

Use ∫tan u du=−ln|cos u|.

Split the fraction: 1/cos²x−1/sin²x=sec²x−cosec²x.

Since (tan x)′=sec²x and (cot x)′=−cosec²x, integrate both terms.

Let u=eˣ; then du=eˣdx.

The integral becomes ∫sec²u du.

Set u=x³, so du=3x²dx.

Then x⁶+1=u²+1, and the integral becomes ∫du/(u²+1).

Factor 4 inside the square root: √(1+4x²)=2√(x²+1/4).

Use the standard form ∫dx/√(x²+a²)=ln|x+√(x²+a²)|+C.

Set u=2−x, so du=−dx.

Use ∫du/√(u²+1)=ln|u+√(u²+1)|+C, remembering dx=−du.

Set u=5x/3, so dx=(3/5)du.

Then √(9−25x²)=3√(1−u²), so the constants leave (1/5)∫du/√(1−u²).

Set u=x², so du=2x dx.

The integral becomes (3/2)∫du/(1+2u²).

Set v=√2u and use the inverse-tangent formula.

Set u=x³, so du=3x²dx.

Then the integral is (1/3)∫du/(1−u²).

Use ∫du/(1−u²)=(1/2)ln|(1+u)/(1−u)|+C.

Split the numerator: (x−1)/√(x²−1)=x/√(x²−1)−1/√(x²−1).

The first term integrates to √(x²−1) by setting u=x²−1.

Use ∫dx/√(x²−1)=ln|x+√(x²−1)|+C for the second term.

Set u=x³, so du=3x²dx.

The integral becomes (1/3)∫du/√(u²+a⁶).

Use ∫du/√(u²+k²)=ln|u+√(u²+k²)|+C.

Set u=tan x, so du=sec²x dx.

The integral becomes ∫du/√(u²+4).

Use the standard form ∫du/√(u²+a²)=ln|u+√(u²+a²)|+C.

Complete the square: x²+2x+2=(x+1)²+1.

Set u=x+1 and apply ∫du/√(u²+1)=ln|u+√(u²+1)|+C.

Complete the square: 9x²+6x+5=(3x+1)²+4.

Set u=3x+1, so dx=du/3.

Use ∫du/(u²+4)=(1/2)tan⁻¹(u/2)+C.

Complete the square: 7−6x−x²=16−(x+3)².

Set u=(x+3)/4. Then dx=4du and the square root is 4√(1−u²).

Complete the square: (x−1)(x−2)=(x−3/2)²−1/4.

Use ∫du/√(u²−a²)=ln|u+√(u²−a²)|+C with u=x−3/2 and a=1/2.

Complete the square: 8+3x−x²=41/4−(x−3/2)².

Set u=(2x−3)/√41. The constants reduce the integral to ∫du/√(1−u²).

Complete the square: (x−a)(x−b)=[x−(a+b)/2]²−[(a−b)/2]².

Use ∫du/√(u²−k²)=ln|u+√(u²−k²)|+C with u=x−(a+b)/2 and k=|a−b|/2.

Let u=2x²+x−3. Then du=(4x+1)dx, which is exactly the numerator times dx.

The integral becomes ∫du/√u.

Use the power rule: ∫u^(−1/2)du=2√u.

Split x+2 into x+2.

For ∫x/√(x²−1)dx, set u=x²−1; this gives √(x²−1).

The remaining term is 2∫dx/√(x²−1)=2ln|x+√(x²−1)|+C.

The denominator D=3x²+2x+1 has derivative D′=6x+2. Write 5x−2=(5/6)D′−11/3.

The first term integrates to (5/6)ln(3x²+2x+1).

Complete the square: 3x²+2x+1=3[(x+1/3)²+2/9]. The remaining integral is an inverse tangent.

Let D=(x−5)(x−4)=x²−9x+20, so D′=2x−9. Rewrite 6x+7=3D′+34.

The first part integrates to 6√D.

Complete the square: D=(x−9/2)²−1/4. Thus ∫dx/√D=ln|x−9/2+√D|+C.

Rewrite 4x−x²=4−(x−2)² and x+2=(x−2)+4.

For ∫(x−2)/√(4−(x−2)²)dx, use u=4−(x−2)²; the result is −√(4x−x²).

For the remaining term use ∫dx/√(4−(x−2)²)=sin⁻¹((x−2)/2).

Let D=x²+2x+3, so D′=2x+2 and x+2=(1/2)D′+1.

The derivative-matching part integrates to √D.

Since D=(x+1)²+2, the remaining standard integral is ln|x+1+√D|+C.

Let D=x²−2x−5, so D′=2x−2 and x+3=(1/2)D′+4.

The derivative-matching part gives (1/2)ln|D|.

Complete the square: D=(x−1)²−6. Use ∫du/(u²−a²)=(1/(2a))ln|(u−a)/(u+a)|+C with u=x−1 and a=√6.

Let D=x²+4x+10, so D′=2x+4 and 5x+3=(5/2)D′−7.

The derivative-matching part integrates to 5√D.

Complete the square: D=(x+2)²+6. Then ∫dx/√D=ln|x+2+√D|+C.

Complete the square: x²+2x+2=(x+1)²+1.

Set u=x+1 and integrate 1/(1+u²).

Complete the square: 9x−4x²=81/16−4(x−9/8)².

Set u=(8x−9)/9. Then dx=(9/8)du and √(9x−4x²)=(9/4)√(1−u²).

The integral becomes (1/2)∫du/√(1−u²).

Write x/[(x+1)(x+2)]=A/(x+1)+B/(x+2).

Matching coefficients gives A=−1 and B=2.

Integrate the two reciprocal terms.

Factor x²−9=(x−3)(x+3).

Use 1/(x²−9)=(1/6)[1/(x−3)−1/(x+3)].

Integrate both terms.

Set the fraction equal to A/(x−1)+B/(x−2)+C/(x−3).

Substituting x=1, 2 and 3 gives A=1, B=−5 and C=4.

Integrate each reciprocal term.

Write the fraction as A/(x−1)+B/(x−2)+C/(x−3).

Substitute x=1, 2 and 3 to find A=1/2, B=−2 and C=3/2.

Integrate the three terms.

Factor x²+3x+2=(x+1)(x+2).

Decompose 2x/[(x+1)(x+2)]=−2/(x+1)+4/(x+2).

Integrate each logarithmic term.

The numerator and denominator have the same degree, so divide first: (1−x²)/[x(1−2x)]=1/2+1/x+(3/2)/(1−2x).

Integrate the constant and reciprocal terms; for the last term set u=1−2x.

Decompose x/[(x²+1)(x−1)]=(1/2)(1−x)/(x²+1)+(1/2)/(x−1).

Integrate the constant-over-quadratic term with tan⁻¹x and the x-over-quadratic term with ln(x²+1).

Integrate the final reciprocal term.

Decompose the fraction as A/(x−1)+B/(x−1)²+C/(x+2).

Clearing denominators and equating coefficients gives A=2/9, B=1/3 and C=−2/9.

Integrate the simple reciprocal and repeated-factor terms.

Factor the denominator: x³−x²−x+1=(x−1)²(x+1).

Partial fractions give −1/[2(x−1)]+4/(x−1)²+1/[2(x+1)].

Integrate the logarithmic and reciprocal-square terms.

Factor x²−1=(x−1)(x+1).

Partial fractions give −1/[10(x−1)]+5/[2(x+1)]−24/[5(2x+3)].

Integrate each reciprocal, using u=2x+3 in the last term.

Factor x²−4=(x−2)(x+2).

Partial fractions give −5/[3(x−1)]+5/[2(x−2)]−5/[6(x+2)].

Integrate each reciprocal term.

Divide first: (x³+x+1)/(x²−1)=x+(2x+1)/(x²−1).

Factor x²−1=(x−1)(x+1) and decompose (2x+1)/(x²−1)=3/[2(x−1)]+1/[2(x+1)].

Integrate the polynomial and reciprocal terms.

Decompose 2/[(1−x)(1+x²)]=1/(1−x)+(x+1)/(1+x²).

Split the second numerator into x and 1.

Integrate the logarithm, the half-logarithm, and tan⁻¹x.

Set u=x+2; then 3x−1=3u−7.

Rewrite the integrand as 3/u−7/u².

Integrate these two powers of u.

Factor x⁴−1=(x−1)(x+1)(x²+1).

Partial fractions give 1/[4(x−1)]−1/[4(x+1)]−1/[2(x²+1)].

Integrate the logarithmic terms and the inverse-tangent term.

Multiply numerator and denominator by xⁿ⁻¹: the fraction becomes xⁿ⁻¹/[xⁿ(xⁿ+1)].

Set t=xⁿ, so dt=nxⁿ⁻¹dx. The integral is (1/n)∫dt/[t(t+1)].

Use 1/[t(t+1)]=1/t−1/(t+1), then integrate.

Set t=sin x, so dt=cos x dx.

Decompose 1/[(1−t)(2−t)]=1/(1−t)−1/(2−t).

Integrate the two logarithmic terms and substitute t=sin x.

Set y=x². The numerator is y²+3y+2 and the denominator is y²+7y+12.

Rewrite the fraction as 1+2/(x²+3)−6/(x²+4).

Use ∫dx/(x²+a²)=(1/a)tan⁻¹(x/a)+C on the last two terms.

Set t=x², so dt=2x dx.

Decompose 1/[(t+1)(t+3)]=(1/2)/(t+1)−(1/2)/(t+3).

Integrate and substitute t=x².

Set t=x⁴ after multiplying numerator and denominator by x³; then x³dx=dt/4.

The integral becomes (1/4)∫dt/[t(t−1)].

Decompose 1/[t(t−1)]=−1/t+1/(t−1) and integrate.

Set t=eˣ, so dx=dt/t.

The integral becomes ∫dt/[t(t−1)].

Use 1/[t(t−1)]=−1/t+1/(t−1), then substitute back.

Partial fractions give x/[(x−1)(x−2)]=−1/(x−1)+2/(x−2).

Integrate the two logarithmic terms and combine them into one logarithm.

Decompose 1/[x(x²+1)]=1/x−x/(x²+1).

Integrate each term.

Use integration by parts with u=x and dv=sin x dx, so du=dx and v=−cos x.

Then ∫x sin x dx=−x cos x+∫cos x dx.

Use integration by parts with u=x and dv=sin 3x dx, so v=−(1/3)cos 3x.

The remaining integral is (1/3)∫cos 3x dx=(1/9)sin 3x.

Use integration by parts with u=x² and dv=eˣdx. This gives ∫x²eˣdx=x²eˣ−2∫xeˣdx.

Apply integration by parts again: ∫xeˣdx=xeˣ−eˣ.

Substitute that result into the first line and factor out eˣ.

Use integration by parts with u=log x and dv=x dx. Then du=dx/x and v=x²/2.

The remaining integral is (1/2)∫x dx=x²/4.

Use integration by parts with u=log(2x) and dv=x dx. Then du=dx/x and v=x²/2.

The remaining integral is (1/2)∫x dx=x²/4.

Use integration by parts with u=log x and dv=x²dx, giving v=x³/3 and du=dx/x.

Then ∫x²log x dx=(x³/3)log x−(1/3)∫x²dx.

Evaluate the remaining power integral.

Use integration by parts with u=sin⁻¹x and dv=x dx. Then du=dx/√(1−x²) and v=x²/2.

The remaining integral is (1/2)∫x²/√(1−x²)dx. Write x²=1−(1−x²), then integrate the two parts.

Use integration by parts with u=tan⁻¹x and dv=x dx. Then du=dx/(1+x²) and v=x²/2.

The remaining integral is (1/2)∫x²/(1+x²)dx=(1/2)∫[1−1/(1+x²)]dx.

Use integration by parts with u=cos⁻¹x and dv=x dx. Then du=−dx/√(1−x²) and v=x²/2.

The remaining integral is +(1/2)∫x²/√(1−x²)dx.

Use x²=1−(1−x²) to evaluate it, as in the preceding inverse-trigonometric integral.

Use integration by parts with u=(sin⁻¹x)² and dv=dx.

Then I=x(sin⁻¹x)²−2∫x sin⁻¹x/√(1−x²)dx.

In the remaining integral, set t=√(1−x²); then x dx=−t dt and sin⁻¹x can be integrated by parts, giving x−√(1−x²)sin⁻¹x.

Notice d(√(1−x²))/dx=−x/√(1−x²).

Use integration by parts with u=cos⁻¹x and dv=−d(√(1−x²)).

The remaining integral is ∫dx=x.

Use integration by parts with u=x and dv=sec²x dx. Then du=dx and v=tan x.

The remaining integral is ∫tan x dx=−log|cos x|.

Write the integral as ∫1·tan⁻¹x dx.

Use integration by parts with u=tan⁻¹x and dv=dx. Then du=dx/(1+x²) and v=x.

The remaining integral is ∫x/(1+x²)dx=(1/2)log(1+x²).

Use integration by parts with u=(log x)² and dv=x dx.

Then I=(x²/2)(log x)²−∫x log x dx.

Use the result of integrating x log x for the remaining integral.

Split the integral into ∫x²log x dx+∫log x dx.

Use integration by parts on each term: ∫x²log x dx=(x³/3)log x−x³/9 and ∫log x dx=xlog x−x.

Differentiate eˣ sin x using the product rule.

d(eˣ sin x)/dx=eˣ sin x+eˣ cos x, exactly the given integrand.

Differentiate eˣ/(1+x) using the quotient rule.

d[eˣ/(1+x)]/dx=eˣ/(1+x)−eˣ/(1+x)²=xeˣ/(1+x)².

Differentiate eˣtan(x/2) using the product rule.

Use d[tan(x/2)]/dx=1/(1+cos x).

Then the derivative is eˣ[tan(x/2)+1/(1+cos x)]=eˣ(1+sin x)/(1+cos x).

Differentiate eˣ/x using the quotient rule.

d(eˣ/x)/dx=eˣ/x−eˣ/x², which is the integrand.

Differentiate eˣ/(x−1)².

Its derivative is eˣ/(x−1)²−2eˣ/(x−1)³=eˣ(x−3)/(x−1)³.

Use integration by parts twice, integrating e^(2x) each time.

Let I=∫e^(2x)sin x dx. The first step gives I=(1/2)e^(2x)sin x−(1/2)∫e^(2x)cos x dx.

Integrating the remaining cosine product by parts gives an equation involving I; solve it for I.

On this interval, sin⁻¹(2x/(1+x²))=2tan⁻¹x because sin(2tan⁻¹x)=2x/(1+x²) and 2tan⁻¹x lies in the principal range of sin⁻¹.

Integrate 2tan⁻¹x by parts, taking u=tan⁻¹x and dv=2dx.

The remaining integral is 2∫x/(1+x²)dx=log(1+x²).

Set u=x³, so du=3x²dx.

Then the integral is (1/3)∫eᵘdu.

Differentiate eˣsec x using the product rule.

d(eˣsec x)/dx=eˣsec x+eˣsec x tan x, which is the given integrand.

This matches ∫√(a²−x²)dx with a=2.

Use ∫√(a²−x²)dx=(x/2)√(a²−x²)+(a²/2)sin⁻¹(x/a)+C.

Set u=2x, so dx=du/2.

Then the integral is (1/2)∫√(1−u²)du.

Apply the standard formula with a=1 and substitute u=2x.

Complete the square: x²+4x+6=(x+2)²+2.

Use ∫√(u²+a²)du=(u/2)√(u²+a²)+(a²/2)ln|u+√(u²+a²)|+C with u=x+2 and a²=2.

Complete the square: x²+4x+1=(x+2)²−3.

Use ∫√(u²−a²)du=(u/2)√(u²−a²)−(a²/2)ln|u+√(u²−a²)|+C with u=x+2 and a²=3.

Complete the square: 1−4x−x²=5−(x+2)².

Use ∫√(a²−u²)du=(u/2)√(a²−u²)+(a²/2)sin⁻¹(u/a)+C with u=x+2 and a²=5.

Complete the square: x²+4x−5=(x+2)²−9.

Apply ∫√(u²−a²)du with u=x+2 and a=3.

Complete the square: 1+3x−x²=13/4−(x−3/2)².

Use ∫√(a²−u²)du=(u/2)√(a²−u²)+(a²/2)sin⁻¹(u/a)+C with u=x−3/2 and a=√13/2, so a²/2=13/8.

Complete the square: x²+3x=(x+3/2)²−9/4.

Apply ∫√(u²−a²)du with u=x+3/2 and a=3/2.

Rewrite the integrand as (1/3)√(x²+9).

Apply ∫√(x²+a²)dx=(x/2)√(x²+a²)+(a²/2)ln|x+√(x²+a²)|+C with a=3, then multiply by 1/3.

Use the standard formula for ∫√(x²+a²)dx with a=1.

Differentiate the resulting expression to check that it returns √(1+x²).

Complete the square: x²−8x+7=(x−4)²−9.

Apply ∫√(u²−a²)du=(u/2)√(u²−a²)−(a²/2)ln|u+√(u²−a²)|+C with u=x−4 and a=3.

Find an antiderivative: ∫(x+1)dx=x²/2+x.

Substitute the upper limit 1 and subtract the value at the lower limit −1.

An antiderivative of 1/x is ln|x|.

Both limits are positive, so evaluate ln 3−ln 2.

Integrate each term: an antiderivative is x⁴−(5/3)x³+3x²+9x.

At x=2 the value is 98/3; at x=1 it is 37/3.

Subtract: 98/3−37/3=61/3.

An antiderivative is −(1/2)cos 2x.

At x=π/4 this is 0; at x=0 it is −1/2.

Subtract: 0−(−1/2)=1/2.

An antiderivative is (1/2)sin 2x.

At x=π/2 it is (1/2)sin π=0; at x=0 it is 0.

An antiderivative of eˣ is eˣ.

Evaluate at the limits: e⁵−e⁴.

Use ∫tan x dx=−ln|cos x|.

At x=π/4, −ln(cos x)=−ln(1/√2)=(1/2)ln 2; at x=0 it is 0.

Use ∫cosec x dx=ln|tan(x/2)|.

Substitute the limits: ln(tan(π/8))−ln(tan(π/12)).

Use tan(π/8)=√2−1 and tan(π/12)=2−√3.

An antiderivative is sin⁻¹x.

Evaluate: sin⁻¹1−sin⁻¹0=π/2−0.

An antiderivative is tan⁻¹x.

Evaluate: tan⁻¹1−tan⁻¹0=π/4−0.

Factor x²−1=(x−1)(x+1) and use partial fractions: 1/(x²−1)=(1/2)[1/(x−1)−1/(x+1)].

An antiderivative is (1/2)ln|(x−1)/(x+1)|.

At the limits the difference is (1/2)ln(1/2)−(1/2)ln(1/3)=(1/2)ln(3/2).

Use cos²x=(1+cos 2x)/2.

An antiderivative is x/2+(1/4)sin 2x.

At π/2 the value is π/4; at 0 it is 0.

The denominator x²+1 has derivative 2x, so an antiderivative is (1/2)ln(x²+1).

Evaluate: (1/2)ln 10−(1/2)ln 5.

Split the numerator: (2x+3)/(5x²+1)=2x/(5x²+1)+3/(5x²+1).

The first integral gives (1/5)ln(5x²+1). For the second, use ∫dx/(1+a²x²)=(1/a)tan⁻¹(ax), with a=√5.

Evaluate from 0 to 1: (1/5)ln 6+(3/√5)tan⁻¹(√5).

Let u=x², so du=2x dx and x dx=du/2.

The limits become u=0 and u=1, giving (1/2)∫₀¹eᵘdu.

Evaluate: (1/2)(e−1).

Factor the denominator: (x+1)(x+3). Divide to get 5x²/(x²+4x+3)=5+5/[2(x+1)]−45/[2(x+3)].

An antiderivative is 5x+(5/2)ln|x+1|−(45/2)ln|x+3|.

Evaluate from 1 to 2: 5+(5/2)ln(3/2)−(45/2)ln(5/4).

Integrate term by term: an antiderivative is 2tan x+x⁴/4+2x.

At π/4 this is 2+π⁴/1024+π/2; at 0 it is 0.

Expand and use sin²u+cos²u=1 and 2sin u cos u=sin 2u. The integrand becomes 1−sin x.

An antiderivative is x+cos x.

Evaluate: (π+cos π)−(0+cos 0)=π−1−1.

Split the integral into 6x/(x²+4)+3/(x²+4).

An antiderivative is 3ln(x²+4)+(3/2)tan⁻¹(x/2).

At x=2 minus x=0 gives 3ln 2+3π/8.

Integrate each term: an antiderivative is eˣ−(4/π)cos(πx/4).

At x=1 minus x=0 gives e−1+(4/π)(1−cos(π/4)).

Use cos(π/4)=√2/2.

An antiderivative is tan⁻¹x.

Evaluate: tan⁻¹(√3)−tan⁻¹1=π/3−π/4=π/12.

Factor 4 from the denominator: 4+9x²=4[1+(3x/2)²].

Let u=3x/2, so dx=(2/3)du. The limits change from 0 and 2/3 to 0 and 1.

The integral is (1/6)∫₀¹du/(1+u²)=(1/6)(π/4)=π/24.

Let u=x²+1, so du=2x dx.

The limits become u=1 and u=2, so the integral is (1/2)∫₁²du/u.

Let u=sin φ, so du=cos φ dφ. Write cos⁵φ dφ=(1−u²)²du.

The limits become u=0 and u=1. Expand: ∫₀¹u^(1/2)(1−2u²+u⁴)du.

Integrate the powers: 2/3−4/7+2/11=64/231.

For 0≤x≤1, sin(2tan⁻¹x)=2x/(1+x²), and 2tan⁻¹x lies between 0 and π/2. Thus sin⁻¹(2x/(1+x²))=2tan⁻¹x.

An antiderivative of 2tan⁻¹x is 2x tan⁻¹x−ln(1+x²).

At 1 this is π/2−ln 2; at 0 it is 0.

Use the suggested substitution t²=x+2. Then x=t²−2 and dx=2t dt. The limits change from x=0,2 to t=√2,2.

The integral becomes ∫ from √2 to 2 2(t⁴−2t²)dt. An antiderivative is (2/5)t⁵−(4/3)t³.

At 2 the value is 32/15; at √2 it is −16√2/15. Subtract the lower value.

Let u=cos x, so du=−sin x dx. The limits change from u=1 to u=0.

Reverse the limits to remove the minus sign: ∫₀¹du/(1+u²).

Evaluate using tan⁻¹u: tan⁻¹1−tan⁻¹0=π/4.

Complete the square: x+4−x²=17/4−(x−1/2)².

Use ∫dy/(a²−y²)=(1/(2a))ln|(a+y)/(a−y)| with y=x−1/2 and a=√17/2.

Evaluating at x=2 and x=0 gives (1/√17)ln[((√17+3)(√17+1))/((√17−3)(√17−1))].

Complete the square: x²+2x+5=(x+1)²+4.

Let u=(x+1)/2, so dx=2du. The limits become 0 and 1.

The integral becomes (1/2)∫₀¹du/(1+u²)=(1/2)tan⁻¹(1).

Notice that d(e^(2x)/(2x))/dx=e^(2x)(1/x−1/(2x²)), exactly the integrand.

Evaluate the antiderivative at the endpoints: e⁴/4−e²/2.

Set u=(1−x²)/x². Since x−x³=x³u, the integrand becomes u^(1/3)/x³ dx.

Differentiate u=x⁻²−1: du=−2x⁻³dx. The limits change from u=8 at x=1/3 to u=0 at x=1.

The integral is (1/2)∫₀⁸u^(1/3)du=(3/8)·8^(4/3)=6.

By the fundamental theorem of calculus, if f(x)=∫₀ˣg(t)dt, then f′(x)=g(x).

Here g(t)=t sin t, so replace t by x.

Use cos²x=(1+cos 2x)/2.

An antiderivative is x/2+sin 2x/4.

Evaluate at π/2 and 0: π/4−0.

Call the integral I. Replace x by π/2−x; sine and cosine swap, so I also equals the integral with √(cos x) in the numerator.

Add these two equal integrals. The numerators then add to the denominator, leaving ∫₀^(π/2)1 dx=π/2.

Thus 2I=π/2.

Call the integral I and replace x by π/2−x. This gives the same integral with sin and cos exchanged.

Add the two forms. Their numerators add to the denominator, so 2I=∫₀^(π/2)1 dx=π/2.

Let I be the integral. Under x→π/2−x, it becomes the same integral with sin⁵x in the numerator.

Add the two versions: the numerators sum to the denominator, giving 2I=π/2.

The expression x+2 changes sign at x=−2. Split the integral there.

From −5 to −2, the graph is a triangle with base 3 and height 3, so its area is 9/2. From −2 to 5, it is a triangle with base 7 and height 7, so its area is 49/2.

Add the two positive areas: 9/2+49/2=29.

The expression x−5 changes sign at x=5. Split the integral at 5.

Each side forms a triangle with base 3 and height 3, so each area is 9/2.

Add them: 9/2+9/2=9.

Set u=1−x. Then x=1−u, and the limits remain 0 and 1 after reversing their order.

The integral becomes ∫₀¹(1−u)uⁿdu=∫₀¹(uⁿ−u^(n+1))du.

Evaluate: 1/(n+1)−1/(n+2)=1/((n+1)(n+2)).

Call the integral I and substitute x=π/4−t. Then tan x=(1−tan t)/(1+tan t), so 1+tan x=2/(1+tan t).

This gives I=∫₀^(π/4)[log 2−log(1+tan t)]dt=(π/4)log 2−I.

Solve 2I=(π/4)log 2.

Let t=2−x. Then x=2−t and dx=−dt; the limits change from t=2 to t=0.

Reverse the limits: ∫₀²(2−t)t^(1/2)dt=∫₀²(2t^(1/2)−t^(3/2))dt.

Evaluate: (4/3)·2^(3/2)−(2/5)·2^(5/2)=16√2/15.

Use sin 2x=2sin x cos x. Then the integrand becomes −log 2+log(sin x/cos x)=−log 2+log tan x.

The symmetry x→π/2−x changes log tan x to −log tan x, so its integral over this interval is 0.

The constant term contributes −(π/2)log 2.

Use sin²x=(1−cos 2x)/2.

An antiderivative is x/2−sin 2x/4.

The cosine term has equal endpoint values; the x term contributes π/2.

Let I be the integral. Since sin(π−x)=sin x, substituting x→π−x gives I=∫₀^π(π−x)/(1+sin x)dx.

Add the two expressions: 2I=π∫₀^πdx/(1+sin x). Rationalise: 1/(1+sin x)=(1−sin x)/cos²x=sec²x−sec x tan x.

An antiderivative is tan x−sec x. Its endpoint limits give ∫₀^πdx/(1+sin x)=2; hence I=π.

The function sin x is odd, so its seventh power is also odd: sin⁷(−x)=−sin⁷x.

The limits are symmetric around 0, so the contributions cancel in pairs.

Cosine has period 2π, so split the interval into two equal periods of length π if useful; more directly, pair x with π−x on [0,π].

cos(π−x)=−cos x, so the integrals over [0,π] cancel. The remaining interval [π,2π] repeats the same zero total.

The denominator has derivative cos x−sin x, which is the negative of the numerator.

An antiderivative is −log(1+sin x cos x).

At both x=0 and x=π/2, sin x cos x=0, so the endpoint difference is 0.

Use 1+cos x=2cos²(x/2), so log(1+cos x)=log 2+2log cos(x/2).

Substitute u=x/2 in the second integral: the result is πlog 2+4∫₀^(π/2)log cos u du.

Use ∫₀^(π/2)log cos u du=−(π/2)log 2.

Call the integral I. Substitute x=a−t; the limits swap and the integrand becomes √(a−t)/(√(a−t)+√t).

Add this form to the original. The paired numerators add to the denominator, so 2I=∫₀ᵃ1dt=a.

The expression x−1 changes sign at x=1. Split the interval there.

From 0 to 1 the area is a triangle with base 1 and height 1, giving 1/2. From 1 to 4 the area is a triangle with base 3 and height 3, giving 9/2.

Let I=∫₀ᵃf(x)g(x)dx. Substitute x=a−t; the interval reverses back to 0 to a, and the symmetry gives I=∫₀ᵃf(x)g(a−x)dx.

Add the two equal expressions: 2I=∫₀ᵃf(x)[g(x)+g(a−x)]dx.

Use g(x)+g(a−x)=4 to get 2I=4∫₀ᵃf(x)dx, so I=2∫₀ᵃf(x)dx.

The terms x³, x cos x and tan⁵x are odd functions, so each integrates to 0 over the symmetric interval.

The constant 1 integrates to the interval length: π/2−(−π/2)=π.

Replace x by π/2−x. Sine and cosine swap, so the logarithm becomes log((4+3cos x)/(4+3sin x)), the negative of the original integrand.

The substitution leaves the interval unchanged, so the integral equals its own negative.

Factor the denominator: x−x³=x(1−x²). Rewrite 1/[x(1−x²)]=1/x+x/(1−x²).

Integrate the first term as ln|x|. For the second, use u=1−x², so it integrates to −(1/2)ln|1−x²|.

Multiply by the conjugate: 1/(√(x+a)+√(x+b))=[√(x+a)−√(x+b)]/(a−b), when a≠b.

Integrate the two square-root terms using ∫√(x+c)dx=(2/3)(x+c)^(3/2).

Let x=a sin²t. Then dx=2a sin t cos t dt and √(ax−x²)=a sin t cos t.

The integrand becomes (2/a)cosec²t dt, whose antiderivative is −(2/a)cot t.

Since cot t=√((a−x)/x), substitute back.

Differentiate F(x)=(x⁴+1)^(1/4)/x using the quotient rule.

F′(x)=−1/[x²(x⁴+1)^(3/4)], so the required integrand is −F′(x).

Combine the terms in the denominator: 1/(1/x²+1/x³)=x³/(x+1).

Divide: x³/(x+1)=x²−x+1−1/(x+1).

Integrate each term using the power rule and logarithm rule.

Use partial fractions: 5x/[(x+1)(x²+9)]=−1/[2(x+1)]+(x+9)/[2(x²+9)].

Integrate the logarithmic terms and use ∫dx/(x²+9)=(1/3)tan⁻¹(x/3).

Write sin x=sin((x−a)+a)=sin(x−a)cos a+cos(x−a)sin a.

Divide by sin(x−a): the integrand becomes cos a+sin a·cot(x−a).

Integrate the constant and cotangent terms.

Since e^(log x)=x, rewrite the quotient as (x⁵−x⁴)/(x³−x²).

Factor numerator and denominator: x⁴(x−1)/[x²(x−1)]=x².

Integrate x².

Let u=sin x, so du=cos x dx.

The integral becomes ∫du/√(4−u²), the standard inverse-sine form.

Factor the numerator: sin⁸x−cos⁸x=(sin⁴x−cos⁴x)(sin⁴x+cos⁴x).

Now sin⁴−cos⁴=(sin²−cos²)(sin²+cos²)=sin²−cos², and sin⁴+cos⁴=1−2sin²cos².

Cancel the common denominator; the integrand is sin²x−cos²x=−cos 2x.

Use tan(x+a)−tan(x+b)=sin(a−b)/[cos(x+a)cos(x+b)].

Rewrite the integrand as [tan(x+a)−tan(x+b)]/sin(a−b).

Integrate each tangent using ∫tan u du=−ln|cos u|.

Let u=x⁴. Then du=4x³dx, so x³dx=du/4.

The integral becomes (1/4)∫du/√(1−u²).

Let u=eˣ, so du=eˣdx.

Use partial fractions: 1/[(u+1)(u+2)]=1/(u+1)−1/(u+2).

Integrate and substitute u=eˣ.

Use partial fractions: 1/[(x²+1)(x²+4)]=(1/3)[1/(x²+1)−1/(x²+4)].

Apply ∫dx/(x²+a²)=(1/a)tan⁻¹(x/a) to each term.

Let u=cos x. Then du=−sin x dx.

The integral becomes −∫u³du.

Let u=x⁴+1. Then du=4x³dx.

The integral becomes (1/4)∫du/u.

Let u=f(ax+b). Then du=a f′(ax+b)dx, so f′(ax+b)dx=du/a.

For n≠−1, integrate uⁿ/a. For n=−1, integrate 1/(au).

Use sin(x+α)=sin x(cos α+cot x sin α). Then the square root in the denominator is sin²x√(cos α+cot x sin α).

Let t=cot x. Since dt=−cosec²x dx, the factors sin²x and dx cancel, leaving −∫dt/√(cos α+t sin α).

Integrate the linear expression under the square root, then substitute t=cot x.

Put t=√x and y=√((1−t)/(1+t)). Solving for t gives t=(1−y²)/(1+y²), and dx=2t dt.

After substitution, the integral is 8∫y²(y²−1)/(1+y²)³dy=8∫dy/(1+y²)−24∫dy/(1+y²)²+16∫dy/(1+y²)³.

Use the standard antiderivatives of 1/(1+y²), 1/(1+y²)² and 1/(1+y²)³, then substitute y and t back.

Use 1+cos 2x=2cos²x and sin 2x=2sin x cos x.

Then (2+sin 2x)/(1+cos 2x)=sec²x+tan x.

By the product rule, d(eˣ tan x)/dx=eˣ(tan x+sec²x).

Decompose into partial fractions: the expression equals −2/(x+1)+1/(x+1)²+3/(x+2).

Integrate each term. The reciprocal-square term has antiderivative −1/(x+1).

Set x=cos 2t, where 0<t<π/2. Then √((1−x)/(1+x))=tan t, so the inverse tangent is t.

Since dx=−2sin 2t dt, the integral becomes ∫−2t sin 2t dt.

Integrate by parts: this is t cos 2t−(1/2)sin 2t+C. Substitute t=(1/2)cos⁻¹x.

Combine the logarithms: log(x²+1)−2log x=log((x²+1)/x²).

Set t=1/x. The integral becomes −∫t√(1+t²)log(1+t²)dt.

Let u=1+t². Integrating −(1/2)u^(1/2)log u gives −(1/3)u^(3/2)log u+(2/9)u^(3/2)+C.

Substitute u=(x²+1)/x².

Use half-angle identities: (1−sin x)/(1−cos x)=(1/2)cosec²(x/2)−cot(x/2).

This is the derivative of −cot(x/2). Therefore an antiderivative of the full integrand is −eˣcot(x/2).

At x=π the expression is 0; at x=π/2 it is −e^(π/2). Subtract lower from upper.

Set u=sin²x. Then du=2sin x cos x dx, and cos⁴x+sin⁴x=1−2u+2u².

The limits become 0 and 1/2, so the integral is (1/2)∫₀^(1/2)du/(1−2u+2u²).

Since d[tan⁻¹(2u−1)]/du=1/(1−2u+2u²), evaluate the endpoints.

Set t=tan x. The integral becomes ∫₀^∞dt/[(1+t²)(1+4t²)].

Decompose: 1/[(1+t²)(1+4t²)]=−1/[3(1+t²)]+4/[3(1+4t²)].

Integrate using inverse tangent and take the limit as t→∞.

Let u=sin x−cos x. Then du=(sin x+cos x)dx and sin 2x=1−u².

The limits become u=(1−√3)/2 and u=(√3−1)/2.

Integrate 1/√(1−u²) as sin⁻¹u and evaluate.

Multiply numerator and denominator by the conjugate √(1+x)+√x. The denominator becomes (1+x)−x=1.

The integrand becomes √(1+x)+√x.

Integrate the two powers and evaluate from 0 to 1: (2/3)(2^(3/2)−1)+(2/3).

Let u=sin x−cos x. Then du=(sin x+cos x)dx and sin 2x=1−u².

The limits become −1 and 0, so the integral is ∫₋₁⁰du/(25−16u²).

An antiderivative is (1/40)ln|(5+4u)/(5−4u)|. Evaluate at 0 and −1.

Let u=sin x, so sin 2x dx=2u du. The limits change from 0 to 1.

Integrate 2u tan⁻¹u by parts: [u²tan⁻¹u]₀¹−∫₀¹u²/(1+u²)du.

Use u²/(1+u²)=1−1/(1+u²).

On [1,4], |x−1|=x−1. Split |x−2| at 2 and |x−3| at 3.

The three areas are 9/2, 5/2 and 5/2, respectively.

Add the contributions.

Decompose: 1/[x²(x+1)]=−1/x+1/x²+1/(x+1).

An antiderivative is −log x−1/x+log(x+1).

Evaluate at 3 and 1.

Use integration by parts with u=x and dv=eˣdx. Then an antiderivative is eˣ(x−1).

Evaluate: e(1−1)−1(0−1).

x¹⁷ is odd and cos⁴x is even. Their product is odd.

An odd function’s positive and negative contributions cancel on limits symmetric about zero.

Write sin³x=sin x(1−cos²x).

Let u=cos x; the limits change from 1 to 0 and the integral becomes ∫₀¹(1−u²)du.

Evaluate [u−u³/3]₀¹.

Use tan³x=tan x(sec²x−1).

An antiderivative of 2tan³x is tan²x+2log|cos x|.

At π/4 this is 1+2log(√2/2)=1−log 2; at 0 it is 0.

Use integration by parts with u=sin⁻¹x and dv=dx.

An antiderivative is x sin⁻¹x+√(1−x²).

Evaluate at 1 and 0.

Multiply numerator and denominator by eˣ to obtain eˣ/(1+e²ˣ).

The derivative of tan⁻¹(eˣ) is eˣ/(1+e²ˣ).

Use cos 2x=(cos x−sin x)(cos x+sin x).

The quotient simplifies to (cos x−sin x)/(sin x+cos x), the derivative of log|sin x+cos x|.

Call the required integral I. Substitute x=a+b−t; the limits reverse and then return to a and b.

The symmetry condition gives I=∫ₐᵇ(a+b−x)f(x)dx.

Add this to the original I: 2I=(a+b)∫ₐᵇf(x)dx.