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.

• Can you actually provide the work you've done so we can see where you might've went wrong? – Crescendo Nov 14 '17 at 4:54
• Just posted a picture. – Molly Taylor Nov 14 '17 at 4:58
• I really like this question! So I went ahead and proposed a few minor improvements (nothing semantically pedantic). If you feel these edits detract from the quality of your thread, then always feel free to roll them back. – gen-z ready to perish Nov 14 '17 at 5:35

\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*}

We have $$\int(u^{1/2}-u^{2+1/2})du=?$$

• I ended up with $\int u^{\frac1 2} - u du$... – Molly Taylor Nov 14 '17 at 4:59
• @Molly, $$u^{1/2}(1-u^2)=?$$ – lab bhattacharjee Nov 14 '17 at 5:00
• Yeah, sorry! And I had a negative outside the integral too. – Molly Taylor Nov 14 '17 at 5:01
• @Molly, u r right! – lab bhattacharjee Nov 14 '17 at 5:04
• Oh... so where did I go wrong then? – Molly Taylor Nov 14 '17 at 5:05

$\int \sin^3x \sqrt {\cos x}\;dx=\int (u^2-1)\sqrt u\;du$ where $u=\cos x.$

Now $(u^2-1)\sqrt u\;=u^{5/2}-u^{1/2}$. This seems to be where you erred. It integrates to $\frac {1}{1+5/2}u^{1+5/2}-\frac {1}{1+1/2}u^{1+1/2}.$