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?

$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \sin ^7 x \cos ^5 x & \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \int_0^{\frac{\pi}{2}} \cos ^5 x \sin ^7 x d x \\
&=\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin ^5 x\cos ^5 x\left(\sin ^2 x+\cos ^2 x\right) d x \\
&=\frac{1}{64} \int_0^{\frac{\pi}{2}} \sin ^5(2 x)d x \\
&=\frac{1}{128} \int_0^{\frac{\pi}{2}}\left(1-\cos ^2 2 x\right)^2 d(\cos 2 x) \\
&=\frac{1}{128}\left[\cos 2 x-\frac{2 \cos ^3 2 x}{3}+\frac{\cos ^5 2 x}{5}\right]_0^{\frac{\pi}{2}} \\
&=\frac{1}{120}
\end{aligned}
$$
