Trigonometric Integrals

$\sin^3 x = \sin^2 x\cdot\sin x$: keep $\sin x\,dx$, let $u = \cos x$.

$\sin^2 x = 1 - \cos^2 x$: $\int(1 - \cos^2 x)\cos^4 x\,\sin x\,dx$.

$du = -\sin x\,dx$: $-\int(u^4 - u^6)\,du = -\dfrac{u^5}{5} + \dfrac{u^7}{7}$.

$-\tfrac15\cos^5 x + \tfrac17\cos^7 x + C$.

Keep $\cos x\,dx$, let $u = \sin x$; the limits become $0$ and $\tfrac12$.

$\cos^2 x = 1 - \sin^2 x$: $\int_0^{\pi/6}\sin^2 x(1 - \sin^2 x)\cos x\,dx$.

$\int_0^{1/2}(u^2 - u^4)\,du = \Big[\dfrac{u^3}{3} - \dfrac{u^5}{5}\Big]_0^{1/2}$.

$\dfrac{1}{24} - \dfrac{1}{160} = \dfrac{17}{480}$.

The $\sin$ power is odd: keep $\sin x\,dx$, let $u = \cos x$.

$\int\dfrac{1 - \cos^2 x}{\sqrt{\cos x}}\,\sin x\,dx$.

$-\int\big(u^{-1/2} - u^{3/2}\big)\,du = -2u^{1/2} + \tfrac25 u^{5/2}$.

$-2\sqrt{\cos x} + \tfrac25\cos^{5/2}x + C$.

$\sin^2(3x) = \tfrac12\big(1 - \cos(6x)\big)$.

$\tfrac12 - \tfrac12\cos(6x)$.

Nothing squared is left.

$\dfrac{x}{2} - \dfrac{1}{12}\sin(6x) + C$.

$\tfrac12(1 - \cos 2x)\cdot\tfrac12(1 + \cos 2x)$.

$\tfrac14\big(1 - \cos^2 2x\big)$.

$\cos^2 2x = \tfrac12(1 + \cos 4x)$, so $\tfrac14\big(\tfrac12 - \tfrac12\cos 4x\big) = \tfrac18 - \tfrac18\cos 4x$.

$\dfrac{x}{8} - \dfrac{1}{32}\sin 4x + C$.

$\sin^4(2x) = \Big(\tfrac12(1 - \cos 4x)\Big)^2$.

$\tfrac14\big(1 - 2\cos 4x + \cos^2 4x\big)$.

$\cos^2 4x = \tfrac12(1 + \cos 8x)$, so $\tfrac38 - \tfrac12\cos 4x + \tfrac18\cos 8x$.

$\Big[\tfrac{3x}{8} - \tfrac18\sin 4x + \tfrac{1}{64}\sin 8x\Big]_0^{\pi/4} = \tfrac{3\pi}{32} - 0 + 0 = \tfrac{3\pi}{32}$.

$\sec^4 x$ is an even power: $u = \tan x$.

$\sec^4 x = \sec^2 x\cdot\sec^2 x$; keep $\sec^2 x\,dx = du$.

$\sec^2 x = 1 + \tan^2 x$: $\int\tan^2 x(1 + \tan^2 x)\sec^2 x\,dx = \int(u^2 + u^4)\,du$.

$\tfrac13\tan^3 x + \tfrac15\tan^5 x + C$.

$\tan^3 x$ is odd and there is a $\sec x$: $u = \sec x$.

Keep $\sec x\tan x\,dx = du$, leaving $\tan^2 x\sec^2 x$.

$\tan^2 x = \sec^2 x - 1$: $\int(u^2 - 1)u^2\,du = \int(u^4 - u^2)\,du$.

$\tfrac15\sec^5 x - \tfrac13\sec^3 x + C$.

Odd $\tan$: $u = \sec x$; the limits become $\sec 0 = 1$ and $\sec\tfrac{\pi}{3} = 2$.

$\tan x\sec^3 x\,dx = \sec^2 x\cdot\sec x\tan x\,dx = u^2\,du$.

No $\tan x$ is left to convert.

$\int_1^2 u^2\,du = \dfrac{8 - 1}{3} = \dfrac73$.

$\tan^7 x = \tan^5 x\sec^2 x - \tan^5 x$.

$u = \tan x$: $\int\tan^5 x\sec^2 x\,dx = \int u^5\,du = \tfrac16\tan^6 x$.

Two more rounds on the leftover: $-\int\tan^5 x\,dx = -\tfrac14\tan^4 x + \int\tan^3 x\,dx = -\tfrac14\tan^4 x + \tfrac12\tan^2 x - \int\tan x\,dx$.

$-\int\tan x\,dx = -\log|\sec x|$. Total: $\tfrac16\tan^6 x - \tfrac14\tan^4 x + \tfrac12\tan^2 x - \log|\sec x| + C$.

$\tan^4 x = \tan^2 x\sec^2 x - \tan^2 x$.

$\int_0^{\pi/4}\tan^2 x\sec^2 x\,dx = \Big[\tfrac13\tan^3 x\Big]_0^{\pi/4} = \tfrac13$.

$\tan^2 x = \sec^2 x - 1$, so $\int_0^{\pi/4}\tan^2 x\,dx = \Big[\tan x - x\Big]_0^{\pi/4} = 1 - \tfrac{\pi}{4}$.

The last piece was $\int 1\,dx = x$. Total: $\tfrac13 - \big(1 - \tfrac{\pi}{4}\big) = \tfrac{\pi}{4} - \tfrac23$.