Evaluate $\int{\sin^{13}x}dx$

Question

Evaluate $\int \sin^{13}x\,dx$.

My solution so far:
$$\int\sin^{13}x\,dx=\int\sin x\sin^{12}x\,dx=\int\sin x(\sin^{2}x)^6\,dx$$

Let $t=\cos x$, then $dt=-\sin x\,dx$.

Now we have: $$\int\sin^{13}x\,dx=-\int(1-t^2)^6,\ dt$$.

How do I proceed from here? I need detailed answer.

Answer 1

Now, you can expand $(1-t^2)^6$ as follows $$(1-t^2)^6=(1-t^2)^3(1-t^2)^3$$ $$=(1-t^6+3t^4-3t^2)(1-t^6+3t^4-3t^2)$$ $$=t^{12}-6t^{10}+15t^{8}-20t^6+15t^4-6t^2+1$$, hence integrate as follows $$-\int (t^{12}-6t^{10}+15t^{8}-20t^6+15t^4-6t^2+1)\ dt$$ $$=-\frac{t^{13}}{13}+\frac{6t^{11}}{11}-\frac{5t^{9}}{3}+\frac{20t^6}{7}-3t^5+2t^3-t+C$$

Answer 2

\begin{align} \int(1-t^2)^6 &= \int \sum_{n=0}^6 {6 \choose n} (-t^2)^n \\ &= \sum_{n=0}^6 {6 \choose n} (-1)^n \int t^{2n} \\ &= \sum_{n=0}^6 {6 \choose n} (-1)^n \left( \frac{t^{2n+1}}{2n+1}\right) + C \\ &= \frac{t^{13}}{13} - \frac{6 t^{11}}{11} + \frac{5t^9}{3} - \frac{20t^7}{7} + 3 t^5 - 2t^3 + t + C \end{align}

Answer 3

An alternative approach is given by De Moivre's formula and the binomial theorem. Since

$$\begin{eqnarray*} \sin^{13}(x) &=& -\frac{i}{2^{13}}\left(e^{ix}-e^{-ix}\right)^{13}\\&=&-\frac{i}{2^{13}}\left(2i\sin(13x)-26i\sin(11 x)+156i\sin(9x)-572i\sin(7x)+1430i\sin(5x)-2574i\sin(3x)+3432i\sin(x)\right)\end{eqnarray*}$$ we have:

$$ \int \sin^{13}(x)\,dx = C-\frac{429 \cos(x)}{1024}+\frac{429\cos(3x)}{4096}-\frac{143\cos(5x)}{4096}+\frac{143\cos(7x)}{14336}-\frac{13\cos(9x)}{6144}+\frac{13\cos(11x)}{45056}-\frac{\cos(13x)}{53248}.$$

Answer 4

Apply Integral Reduction:$$\quad \int \:\sin ^n\left(x\right)dx=-\frac{\cos \left(x\right)\sin ^{n-1}\left(x\right)}{n}+\frac{n-1}{n}\int \sin ^{n-2}\left(x\right)dx$$

$$\color{red}{\int \sin ^{13}\left(x\right)dx=-\frac{\cos \left(x\right)\sin ^{12}\left(x\right)}{13}+\frac{12}{13}\int \sin ^{11}\left(x\right)dx}$$