<s>[INST] Integrate $\int \sin^3(x)\sqrt{\cos(x)} dx$ I tried to solve $\int \sin^3(x)\sqrt{\cos(x)}\,dx$ by setting it equal to 
$$\int \cos^{1/2}(x)\left(1-\cos^2x\right)\sin(x)\,dx $$
and then making $u=\cos(x)$ and $du=-\sin(x)$. I ended up with
$$\frac{-2\cos^{3/2}(x)}{3} + \frac{\cos^2(x)}{2} + C$$
but the book's answer is
$$\frac{2\cos^{7/2}(x)}{7} + \frac{2\cos^{3/2}(x)}{3} + C$$
Could you give a hint as to what I'm doing wrong? Here's my full work.
 [/INST] $$\begin{align*}
\int \sin^3 x \sqrt{\cos x} \, dx
&= \int (1 - \cos^2 x) \cos^{1/2} x \sin x \, dx \\
&= \int (\cos^{1/2} x - \cos^{5/2} x) \sin x \, dx \\
&= \int -(u^{1/2} - u^{5/2}) \, du \\
&= \frac{2}{7} u^{7/2} - \frac{2}{3} u^{3/2} + C \\
&= \frac{2}{7} \cos^{7/2} x - \frac{2}{3} \cos^{3/2} x + C. \end{align*}$$
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