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?

Your expression is correct. Using the double-angle identity $\cos 2x=1-2\sin^2 x$, you can verify that the more awkward Alpha version is also correct. For $7(1-2\sin^2 x)+11=18-14\sin^2 x$ and $\frac{18}{126}=\frac{1}{7}$ and $\frac{14}{126}=\frac{1}{9}$. 
Remark: Fairly often, with trigonometric functions, verification can be more complicated. As a simple example, suppose one calculation gives $2\cos^2 x+C$ as the integral, and another calculation gives $\cos 2x+C$. The functions $\cos 2x$ and $2\cos^2 x$ are not the same, but they differ by a constant, so if one is valid, then so is the other. 
