Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$
$$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$
Now, $\cos xdx = d(\sin x)$
$$\int u^6du - \int u^8du = \frac{1}{7}u^7 - \frac{1}{9}u^9 + C$$
$$\frac{1}{7}\sin^7(x) - \frac{1}{9}\sin^9(x) + C$$
However, WolframAlpha says it's: 
Can anyone tell if those expressions are equal?

$11 + 7 \cos 2x  = 11 + 7 - 14\sin^2 x = 2(9 - 7\cos^2 x) \Rightarrow\\
\dfrac {1}{126} (11 + 7 \cos 2x) = \dfrac{1}{63}(9 - 7\sin^2x) = \boxed{\dfrac{1}{7} - \dfrac{\sin^2x}{9}}\Rightarrow\\
\\
\text{The expressions are equal.}$
