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?

$$\int_0^\frac\pi2\sin^7x\cos^5xdx=\int_0^\frac\pi2\sin^7x\cos^4x\cos xdx=\int_0^\frac\pi2\sin^7x(1-\sin^2x)^2\cos xdx$$
$$=\int_0^1 u^7(1-u^2)^2 du (\text{ Putting }\sin x=u)$$
$$=\int_0^1 (u^7-2u^9+u^{11}) du$$
$$=\left(\frac{u^8}8-2\frac{u^{10}}{10}+\frac{u^{12}}{12}\right)_0^1$$
$$=\frac18-\frac15+\frac1{12}$$
$$=\frac{15-24+10}{120}=\frac1{120} $$
