Integral $\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $ im asked to find the limited integral here but unfortunately im floundering can someone please point me in the right direction?
$$\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $$
step 1 brake up sin and cos so that i can use substitution
$$\int_0^\frac{\pi}{2} \sin^7(x) \cos^4(x) \cos(x) \, dx$$
step 2 apply trig identity
 $$\int_0^\frac{\pi}{2} \sin^7x\ (1-\sin^2 x)^2 \, dx$$
step 3 use $u$-substitution
$$ \text{let}\,\,\, u= \sin(x)\ du=\cos(x)   $$
step 4 apply use substitution
$$\int_0^\frac{\pi}{2} u^7 (1-u^2)^2 du $$
step 5 expand and distribute and change limits of integration
     $$\int_0^1 u^7-2u^9+u^{11}\ du $$ 
step 6 integrate
$$(1^7-2(1)^9+1^{11})-0$$
i would just end up with $1$ however the book answer is $$\frac {1}{120}$$
how can i be so far off?
 A: You forgot to integrate between step $(5)$ and $(6)$! $$\int_0^1 \left(u^7-2u^9+u^{11}\right)\ du \quad =\quad \left(\frac{u^8}{8}  -\frac{u^{10}}{5}   + \frac{u^{12}}{12}\right)\Big|_0^1 = \frac 18 - \frac 15 +\frac 1{12} = \frac 1{120}$$
You're work was fine, otherwise (you left out $\,dx$ from your earlier integrals, and the factor $\cos x$, which turns out to be $\,du$ and so accommodated in the substitution in the second step), but I think your primary lapse was simply forgetting to integrate before evaluating ;-)
A: $$\int_0^\frac\pi2\sin^7x\cos^5xdx=\int_0^\frac\pi2\sin^7x\cos^4x\cos xdx=\int_0^\frac\pi2\sin^7x(1-\sin^2x)^2\cos xdx$$
$$=\int_0^1 u^7(1-u^2)^2 du (\text{ Putting }\sin x=u)$$
$$=\int_0^1 (u^7-2u^9+u^{11}) du$$
$$=\left(\frac{u^8}8-2\frac{u^{10}}{10}+\frac{u^{12}}{12}\right)_0^1$$
$$=\frac18-\frac15+\frac1{12}$$
$$=\frac{15-24+10}{120}=\frac1{120} $$
A: Another approach using $$\frac{m! n!}{(m+n+1)!}=\operatorname{B}(m+1,n+1)=2\int_0^{\frac{\pi}{2}}\cos^{2m+1}x\sin^{2n+1}x \, dx.$$ 
In our case $$\int_0^\frac{\pi}{2} \sin^7x \cos^5x \, dx=\frac{\operatorname{B}(3,4)}{2}=\frac{1}{2}\frac{2! 3!}{6!}=\frac{1}{120}.$$
Here $\operatorname{B}$ denotes Beta function.
A: step 6 is wrong, check it again, it should be $$\frac{u^8}{8} + \frac{u^{12}}{12} - \frac{u^{10} }{5}$$
A: I'm going to show you a general method for integrals like this.
Consider the integral
$$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$
$t=\sin^2x$:
$\therefore dt=2\sin x\cos x\ dx\\\therefore dx=\frac12t^{-1/2}(1-t)^{-1/2}dt\\\therefore x=0\mapsto t=0\\\therefore x=\pi/2\mapsto t=1$
$$I(a,b)=\int_0^1 t^{a/2}(1-t)^{b/2}\frac12t^{-1/2}(1-t)^{-1/2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$
Note the definition of the Beta function:
$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(b,a)$$
Thus
$$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}=I(b,a)$$
Where $\Gamma(x)$ is the Gamma Function.
A: $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \sin ^7 x \cos ^5 x & \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \int_0^{\frac{\pi}{2}} \cos ^5 x \sin ^7 x d x \\
&=\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin ^5 x\cos ^5 x\left(\sin ^2 x+\cos ^2 x\right) d x \\
&=\frac{1}{64} \int_0^{\frac{\pi}{2}} \sin ^5(2 x)d x \\
&=\frac{1}{128} \int_0^{\frac{\pi}{2}}\left(1-\cos ^2 2 x\right)^2 d(\cos 2 x) \\
&=\frac{1}{128}\left[\cos 2 x-\frac{2 \cos ^3 2 x}{3}+\frac{\cos ^5 2 x}{5}\right]_0^{\frac{\pi}{2}} \\
&=\frac{1}{120}
\end{aligned}
$$
