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MATH
[1251]

Examples from Trigonometric integrals

All examples from Trigonometric integrals.

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∫cos⁡4xsin⁡3x dx\int \cos^4 x \sin^3 x \, dx

Here mm is even and nn is odd. Peel sin⁡x\sin x and replace sin⁡2x=1−cos⁡2x\sin^2 x=1-\cos^2 x:

∫cos⁡4xsin⁡3x dx=∫cos⁡4xsin⁡2xsin⁡x dx=∫cos⁡4x(1−cos⁡2x)sin⁡x dx=∫(cos⁡4x−cos⁡6x)sin⁡x dx.\begin{aligned} \int \cos^4 x \sin^3 x \, dx &=\int \cos^4 x \sin^2 x \sin x \, dx \\ &=\int \cos^4 x (1-\cos^2 x) \sin x \, dx \\ &=\int (\cos^4 x-\cos^6 x) \sin x \, dx. \end{aligned}

Let u=cos⁡xu=\cos x, so du=−sin⁡x dxdu=-\sin x \, dx. The integral is

−∫(u4−u6) du=−u55+u77+C=−cos⁡5x5+cos⁡7x7+C.-\int (u^4-u^6) \, du=-\frac{u^5}{5}+\frac{u^7}{7}+C=-\frac{\cos^5 x}{5}+\frac{\cos^7 x}{7}+C.

[routine]

  1. [1]:Odd power of sine or cosine

∫cos⁡5xsin⁡4x dx\int \cos^5 x \sin^4 x \, dx

Here mm is odd and nn is even. Peel cos⁡x\cos x and replace cos⁡2x=1−sin⁡2x\cos^2 x=1-\sin^2 x:

∫cos⁡5xsin⁡4x dx=∫cos⁡4xsin⁡4xcos⁡x dx=∫(1−sin⁡2x)2sin⁡4xcos⁡x dx=∫(sin⁡4x−2sin⁡6x+sin⁡8x)cos⁡x dx.\begin{aligned} \int \cos^5 x \sin^4 x \, dx &=\int \cos^4 x \sin^4 x \cos x \, dx \\ &=\int (1-\sin^2 x)^2 \sin^4 x \cos x \, dx \\ &=\int \bigl(\sin^4 x-2\sin^6 x+\sin^8 x\bigr) \cos x \, dx. \end{aligned}

Let u=sin⁡xu=\sin x, so du=cos⁡x dxdu=\cos x \, dx. The integral is

∫(u4−2u6+u8) du=u55−2u77+u99+C=sin⁡5x5−2sin⁡7x7+sin⁡9x9+C.\int (u^4-2u^6+u^8) \, du=\frac{u^5}{5}-\frac{2u^7}{7}+\frac{u^9}{9}+C=\frac{\sin^5 x}{5}-\frac{2\sin^7 x}{7}+\frac{\sin^9 x}{9}+C.

[exam]

  1. [1]:Odd power of sine or cosine
  2. [2]:∫0π/2cos⁡5xsin⁡4x dx\int_0^{\pi/2}\cos^5 x\sin^4 x \, dx

∫cos⁡5xsin⁡5x dx\int \cos^5 x \sin^5 x \, dx

Both mm and nn are odd. Factoring out cos⁡x\cos x and substituting u=sin⁡xu=\sin x (so du=cos⁡x dxdu=\cos x \, dx),

∫cos⁡5xsin⁡5x dx=∫cos⁡4xsin⁡5xcos⁡x dx=∫(1−sin⁡2x)2sin⁡5xcos⁡x dx=∫(sin⁡5x−2sin⁡7x+sin⁡9x)cos⁡x dx=∫(u5−2u7+u9) du=u66−2u88+u1010+C=sin⁡6x6−sin⁡8x4+sin⁡10x10+C.\begin{aligned} \int \cos^5 x \sin^5 x \, dx &=\int \cos^4 x \sin^5 x \cos x \, dx \\ &=\int (1-\sin^2 x)^2 \sin^5 x \cos x \, dx \\ &=\int \bigl(\sin^5 x-2\sin^7 x+\sin^9 x\bigr) \cos x \, dx \\ &=\int (u^5-2u^7+u^9) \, du \\ &=\frac{u^6}{6}-\frac{2u^8}{8}+\frac{u^{10}}{10}+C \\ &=\frac{\sin^6 x}{6}-\frac{\sin^8 x}{4}+\frac{\sin^{10} x}{10}+C. \end{aligned}

Factoring out sin⁡x\sin x instead produces a different-looking formula. Both formulas have derivative cos⁡5xsin⁡5x\cos^5x\sin^5x, so on any interval they differ by a constant.

[exam]

  1. [1]:Odd power of sine or cosine

∫cos⁡2x dx\int \cos^2 x \, dx and ∫sin⁡2x dx\int \sin^2 x \, dx

When one of mm or nn is zero the identities are the whole calculation:

∫cos⁡2x dx=∫1+cos⁡2x2 dx=x2+sin⁡2x4+C,\int \cos^2 x \, dx=\int \frac{1+\cos 2x}{2} \, dx=\frac{x}{2}+\frac{\sin 2x}{4}+C,∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C.\int \sin^2 x \, dx=\int \frac{1-\cos 2x}{2} \, dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.

[routine]

  1. [1]:Both powers even
  2. [2]:A repeated irreducible quadratic

∫cos⁡2xsin⁡4x dx\int \cos^2 x \sin^4 x \, dx

Both powers are even, so

∫cos⁡2xsin⁡4x dx=∫1+cos⁡2x2(1−cos⁡2x2)2 dx=18∫(1+cos⁡2x)(1−cos⁡2x)2 dx.\int \cos^2 x \sin^4 x \, dx=\int \frac{1+\cos 2x}{2}\left(\frac{1-\cos 2x}{2}\right)^2 \, dx=\frac{1}{8}\int (1+\cos 2x)(1-\cos 2x)^2 \, dx.

Expanding and cancelling −cos⁡2x-\cos 2x against +cos⁡2x+\cos 2x gives

(1+cos⁡2x)(1−2cos⁡2x+cos⁡22x)=1−cos⁡2x−cos⁡22x+cos⁡32x,(1+\cos 2x)(1-2\cos 2x+\cos^2 2x)=1-\cos 2x-\cos^2 2x+\cos^3 2x,

so the integral is

18(∫1 dx−∫cos⁡2x dx−∫cos⁡22x dx+∫cos⁡32x dx).\frac{1}{8}\left(\int 1 \, dx-\int\cos 2x \, dx-\int\cos^2 2x \, dx+\int\cos^3 2x \, dx\right).

The first two integrals evaluate to xx and 12sin⁡2x\frac{1}{2}\sin 2x. For the even square, cos⁡22x=1+cos⁡4x2\cos^2 2x=\frac{1+\cos 4x}{2}. For the odd cube, cos⁡32x=(1−sin⁡22x)cos⁡2x\cos^3 2x=(1-\sin^2 2x)\cos 2x. Thus the integral equals

18[x−sin⁡2x2−∫1+cos⁡4x2 dx+∫(1−sin⁡22x)cos⁡2x dx].\frac{1}{8}\left[x-\frac{\sin 2x}{2}-\int\frac{1+\cos 4x}{2} \, dx+\int(1-\sin^2 2x)\cos 2x \, dx\right].

The remaining cosine-square integral is x2+sin⁡4x8\frac{x}{2}+\frac{\sin 4x}{8}. For the last integral, let u=sin⁡2xu=\sin 2x, so du=2cos⁡2x dxdu=2\cos 2x \, dx and cos⁡2x dx=12 du\cos 2x \, dx=\tfrac12 \, du. Collecting factors,

x8−sin⁡2x16−x16−sin⁡4x64+116∫(1−u2) du=x16−sin⁡2x16−sin⁡4x64+116(u−u33)+C.\frac{x}{8}-\frac{\sin 2x}{16}-\frac{x}{16}-\frac{\sin 4x}{64}+\frac{1}{16}\int (1-u^2) \, du =\frac{x}{16}-\frac{\sin 2x}{16}-\frac{\sin 4x}{64}+\frac{1}{16}\left(u-\frac{u^3}{3}\right)+C.

Back-substituting u=sin⁡2xu=\sin 2x yields

x16−sin⁡2x16−sin⁡4x64+sin⁡2x16−sin⁡32x48+C=x16−sin⁡4x64−sin⁡32x48+C.\frac{x}{16}-\frac{\sin 2x}{16}-\frac{\sin 4x}{64}+\frac{\sin 2x}{16}-\frac{\sin^3 2x}{48}+C=\frac{x}{16}-\frac{\sin 4x}{64}-\frac{\sin^3 2x}{48}+C.

[exam]

  1. [1]:Both powers even

∫sin⁡3xsin⁡2x dx\int \sin 3x \sin 2x \, dx

With A=3xA=3x and B=2xB=2x,

∫sin⁡3xsin⁡2x dx=−12∫(cos⁡5x−cos⁡x) dx=−12(sin⁡5x5−sin⁡x)+C=−sin⁡5x10+sin⁡x2+C.\begin{aligned} \int \sin 3x \sin 2x \, dx &=-\frac{1}{2}\int \bigl(\cos 5x-\cos x\bigr) \, dx \\ &=-\frac{1}{2}\left(\frac{\sin 5x}{5}-\sin x\right)+C \\ &=-\frac{\sin 5x}{10}+\frac{\sin x}{2}+C. \end{aligned}

[exam]

  1. [1]:Product-to-sum

∫tan⁡x dx\int \tan x \, dx and ∫sec⁡x dx\int \sec x \, dx

Writing tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x},

∫tan⁡x dx=∫sin⁡xcos⁡x dx=−ln⁡∣cos⁡x∣+C.\int \tan x \, dx=\int \frac{\sin x}{\cos x} \, dx=-\ln|\cos x|+C.

For the secant integral, multiply numerator and denominator by sec⁡x+tan⁡x\sec x+\tan x:

∫sec⁡x dx=∫sec⁡x(sec⁡x+tan⁡x)sec⁡x+tan⁡x dx=∫sec⁡2x+sec⁡xtan⁡xsec⁡x+tan⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C,\int \sec x \, dx=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \, dx=\int \frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x} \, dx=\ln|\sec x+\tan x|+C,

since the numerator is then the derivative of the denominator.

[routine]

  1. [1]:Tangent and secant

∫sec⁡3x dx\int \sec^3 x \, dx

Write I=∫sec⁡3x dxI=\int \sec^3 x \, dx. Split the integrand as sec⁡x⋅sec⁡2x\sec x\cdot\sec^2 x and integrate by parts with u=sec⁡xu=\sec x and dv=sec⁡2x dxdv=\sec^2 x \, dx, so du=sec⁡xtan⁡x dxdu=\sec x\tan x \, dx and v=tan⁡xv=\tan x. Then

I=sec⁡xtan⁡x−∫sec⁡xtan⁡2x dx.I=\sec x\tan x-\int \sec x\tan^2 x \, dx.

Replace tan⁡2x=sec⁡2x−1\tan^2 x=\sec^2 x-1:

I=sec⁡xtan⁡x−∫sec⁡x(sec⁡2x−1) dx=sec⁡xtan⁡x−∫sec⁡3x dx+∫sec⁡x dx.I=\sec x\tan x-\int \sec x(\sec^2 x-1) \, dx=\sec x\tan x-\int \sec^3 x \, dx+\int \sec x \, dx.

The first integral on the right is II again, so

2I=sec⁡xtan⁡x+∫sec⁡x dx=sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣+C.2I=\sec x\tan x+\int \sec x \, dx=\sec x\tan x+\ln|\sec x+\tan x|+C.

After dividing by 22 (and absorbing the factor into the arbitrary constant),

I=12sec⁡xtan⁡x+12ln⁡∣sec⁡x+tan⁡x∣+C.I=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C.

[exam]

  1. [1]:Powers of tangent and secant
  2. [2]:Secant powers on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

∫sec⁡4xtan⁡2x dx\int \sec^4 x \tan^2 x \, dx

The power of sec⁡\sec is even, so factor out sec⁡2x\sec^2 x:

∫sec⁡4xtan⁡2x dx=∫sec⁡2xtan⁡2xsec⁡2x dx.\int \sec^4 x \tan^2 x \, dx=\int \sec^2 x \tan^2 x \sec^2 x \, dx.

Let u=tan⁡xu=\tan x, so du=sec⁡2x dxdu=\sec^2 x \, dx. The remaining sec⁡2x\sec^2 x is 1+tan⁡2x1+\tan^2 x and the integral becomes

∫(1+tan⁡2x)tan⁡2xsec⁡2x dx=∫(1+u2)u2 du=∫(u2+u4) du=u33+u55+C=tan⁡3x3+tan⁡5x5+C.\int (1+\tan^2 x)\tan^2 x \sec^2 x \, dx=\int (1+u^2)u^2 \, du=\int (u^2+u^4) \, du=\frac{u^3}{3}+\frac{u^5}{5}+C=\frac{\tan^3 x}{3}+\frac{\tan^5 x}{5}+C.

[exam]

  1. [1]:Powers of tangent and secant

∫sec⁡xtan⁡3x dx\int \sec x \tan^3 x \, dx

Here mm and nn are both odd, so factor out tan⁡xsec⁡x\tan x\sec x:

∫sec⁡xtan⁡3x dx=∫tan⁡2x (tan⁡xsec⁡x) dx.\int \sec x \tan^3 x \, dx=\int \tan^2 x\,(\tan x\sec x) \, dx.

Let u=sec⁡xu=\sec x, so du=sec⁡xtan⁡x dxdu=\sec x\tan x \, dx. Replace tan⁡2x=sec⁡2x−1\tan^2 x=\sec^2 x-1:

∫(sec⁡2x−1)tan⁡xsec⁡x dx=∫(u2−1) du=u33−u+C=sec⁡3x3−sec⁡x+C.\int (\sec^2 x-1)\tan x\sec x \, dx=\int (u^2-1) \, du=\frac{u^3}{3}-u+C=\frac{\sec^3 x}{3}-\sec x+C.

[exam]

  1. [1]:Powers of tangent and secant

∫sec⁡xtan⁡2x dx\int \sec x \tan^2 x \, dx

Here nn is odd and mm is even. Replace tan⁡2x=sec⁡2x−1\tan^2 x=\sec^2 x-1:

∫sec⁡xtan⁡2x dx=∫sec⁡x(sec⁡2x−1) dx=∫sec⁡3x dx−∫sec⁡x dx.\int \sec x \tan^2 x \, dx=\int \sec x(\sec^2 x-1) \, dx=\int \sec^3 x \, dx-\int \sec x \, dx.

The antiderivatives of sec⁡x\sec x and sec⁡3x\sec^3 x found above give

12sec⁡xtan⁡x+12ln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x+tan⁡x∣+C=12sec⁡xtan⁡x−12ln⁡∣sec⁡x+tan⁡x∣+C.\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|-\ln|\sec x+\tan x|+C=\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln|\sec x+\tan x|+C.

[exam]

  1. [1]:Powers of tangent and secant

On this page

∫cos⁡4xsin⁡3x dx\int \cos^4 x \sin^3 x \, dx∫cos⁡5xsin⁡4x dx\int \cos^5 x \sin^4 x \, dx∫cos⁡5xsin⁡5x dx\int \cos^5 x \sin^5 x \, dx∫cos⁡2x dx\int \cos^2 x \, dx and ∫sin⁡2x dx\int \sin^2 x \, dx∫cos⁡2xsin⁡4x dx\int \cos^2 x \sin^4 x \, dx∫sin⁡3xsin⁡2x dx\int \sin 3x \sin 2x \, dx∫tan⁡x dx\int \tan x \, dx and ∫sec⁡x dx\int \sec x \, dx∫sec⁡3x dx\int \sec^3 x \, dx∫sec⁡4xtan⁡2x dx\int \sec^4 x \tan^2 x \, dx∫sec⁡xtan⁡3x dx\int \sec x \tan^3 x \, dx∫sec⁡xtan⁡2x dx\int \sec x \tan^2 x \, dx