I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2} (\sin x)^7 (\cos x)^5 \mathrm{d} x$$
I tried rewriting as
$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\sin x(1-\ (\cos x)^3)\cdot\ (\cos x)^5 \mathrm{d} x$$
but that seems to only give me a very, very long loop that doesn't help me at all. How do I proceed?
$$\int_0^{\pi/2} \sin x\cdot (\sin x)^6\cdot (\cos x)^5 \mathrm{d} x = \sin x(1- (\cos x)^3)\cdot (\cos x)^5 \mathrm{d} x$$
$u = \cos x$, then $du = -\sin xdx$
$\int \frac{-u^6}{6} \mathrm{d} u – \int \frac{-u^9}{9} \mathrm{d} u$
From here it looks like I have an incredibly long string of $u$ substitutions to make to get to something I can find an antiderivative for.
Best Answer
To integrate the function $$f(x)=\sin ^{n}x\cdot\cos ^{m}x,$$ when $n$ or $m$ are positive odd numbers, we can apply a general technique which consists of expanding $f(x)$ into a sum of terms of the form $$\sin ^{p}x\cdot \cos x,\qquad p=1,2,\ldots $$ or $$\cos ^{q}x\cdot \sin x,\qquad q=1,2,\ldots.$$ Using the identity $$\begin{equation*} \cos ^{2}x=1-\sin ^{2}x, \end{equation*}$$ in the form $$\begin{equation*} \cos ^{4}x=(1-\sin ^{2}x)^2 \end{equation*}=1-2\sin ^{2}x+\sin ^{4}x,$$
we rewrite our $$f(x)=\sin ^{7}x\cdot\cos ^{5}x=\sin ^{7}x\cdot\cos ^{4}x\cdot\cos x$$ as $$\begin{eqnarray*} f(x) &=&\sin ^{7}x\cdot \left( 1-2\sin ^{2}x+\sin ^{4}x\right) \cdot \cos x \\ &=&\sin ^{7}x\cdot \cos x-2\sin ^{9}x\cdot \cos x+\sin ^{11}x\cdot \cos x. \end{eqnarray*}$$
Each term is of the form $\sin ^{p}x\cdot \cos x$ and can easily be integrated by the substitution $u=\sin x$, $u^{\prime }=\cos x$, $du=\cos x\;dx=u'\;dx$: $$\begin{eqnarray*} \int \sin ^{p}x\cdot \cos x\;dx &=&\int u^{p}\;du=\frac{u^{p+1}}{p+1}=\frac{\sin ^{p+1}x}{p+1}+C, \\ \int_{0}^{\pi /2}\sin ^{p}x\cdot \cos x\;dx &=&\frac{1}{p+1}. \end{eqnarray*}$$
Added: detailed computation in view of OP's comment
$$\begin{eqnarray*} \int_{0}^{\pi /2}f(x)dx &=&\int_{0}^{\pi /2}\sin ^{7}x\cos ^{5}xdx \\ &=&\int_{0}^{\pi /2}\sin ^{7}x\cdot \cos xdx-2\int_{0}^{\pi /2}\sin ^{9}x\cdot \cos xdx \\ &&+\int_{0}^{\pi /2}\sin ^{11}x\cdot \cos xdx \\ &=&\frac{1}{8}-2\cdot \frac{1}{10}+\frac{1}{12}=\frac{1}{120}. \end{eqnarray*}$$