$$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$ $$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$ 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.
$11 + 7 \cos 2x = 11 + 7 - 14\sin^2 x = 2(9 - 7\cos^2 x) \Rightarrow\\ \dfrac {1}{126} (11 + 7 \cos 2x) = \dfrac{1}{63}(9 - 7\sin^2x) = \boxed{\dfrac{1}{7} - \dfrac{\sin^2x}{9}}\Rightarrow\\ \\ \text{The expressions are equal.}$
