How do I solve $\int\sin^5 x$?

Question

Should I transform $\int\sin^5 x$ into $\int (\sin^2 x)^2 \sin x \; \Bbb d x$ to solve it? Or should I use a different trigonometric identity?

Answer 1

Here, you can make the change of variable $u=\cos(x)$ and use $\sin^2(x)=1-\cos^2(x)$. Then, you just have a polynomial to integrate.

Answer 2

$$\int \sin^5x dx = \int \sin x(\sin^2x)^2 dx = \int \sin x(1-\cos^2x)^2 dx $$

Letting $u = \cos x \implies du = -\sin(x)dx$ we get

$$\int (1-\cos^2x)^2 \sin x dx = -\int (1-u^2)^2 du = \int -1+2u^2 -u^4 du = -u + \dfrac{2}{3} u^3 - \dfrac{u^5}{5} + C = - \cos x + \dfrac{2}{3} \cos ^3 x - \dfrac{\cos^5x}{5} + C$$

Answer 3

$$\int (\sin^2 x)^2\sin x dx=\int (1-\cos^2 x)^2\sin x dx=\int (1-2\cos^2 x+\cos^4 x)\sin x dx$$ $$=-\cos x+\frac{2}{3}\cos^3 x-\frac{1}{5}\cos^5 x+C$$

Answer 4

$$\sin^5 x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^5 = \frac{1}{32i}\left(e^{5ix}-5e^{3ix}+10 e^{ix}-10 e^{-ix}+5e^{-3ix}-e^{-5ix}\right)$$ gives: $$ \sin^5 x = \frac{1}{16}\left(\sin(5x)-5\sin(3x)+10\sin(x)\right)$$ that is way easier to integrate.

Answer 5

Another answer from a totally different point of view.

I found that:

$$ \int \sin^q{x} \, \mathrm{d} x = -\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-q}{2};\frac{3}{2};\cos ^2(x)\right),$$ which leads to $-\frac{1}{15} \cos (x) \left(3 \cos ^4(x)-10 \cos ^2(x)+15\right)$ for $q = 5$.

Cheers!