Solving $\int \sin^3 2 \theta \sqrt{\cos 2 \theta}\:d\theta$ using Integration by Parts. We can solve the following problem by using substitution and trigonometric identities. Such is the solution: $$
 \begin{aligned}
& \int \sin ^3 2 \theta \sqrt{\cos 2 \theta} d \theta \\
=& \int \sin ^3 2 \theta(\cos 2 \theta)^{\frac{1}{2}} \\
=& \int \sin ^2 2 \theta \cdot \sin 2 \theta(\cos 2 \theta)^{1 / 2} \\
=& \int\left(1-\cos ^2 2 \theta\right)(\cos 2 \theta)^{1 / 2} \sin 2 \theta d \theta \\
\text { let } & u=\cos 2 \theta \\
& d u=-2 \sin 2 \theta d \theta \\
& d \theta=-\frac{d u}{2 \sin 2 \theta} \\
=& \int\left(1-u^2\right)(u)^{1 / 2} \cdot \sin 2 \theta \cdot \frac{d u}{2 \sin 2 \theta} \\
=&-\frac{1}{2} \int\left(1-u^2\right)(u)^{1 / 2} d u \\
=&-\frac{1}{2} \int\left(u^{1 / 2}-u^{3 / 2}\right) d u \\
&=-\frac{1}{2}\left[\frac{2}{3} u^{3 / 2}-\frac{2}{5} u^{5 / 2}\right] \\
&=-\frac{1}{3} u^{3 / 2}+\frac{1}{5} u^{5 / 2}+C \\
&=-\frac{1}{3} \cos ^{3 / 2} 2 \theta+\frac{1}{5} \cos ^{5 / 2} 2 \theta+C
\end{aligned}$$
We are trying to find out how to solve the problem using Integration by Parts (The one where it says $\int udv = uv - \int vdu$). We have arrived at the solution below but we do not know how to continue from here until we get the desired result which is $-\frac{1}{3} \cos ^{3 / 2} 2 \theta+\frac{1}{5} \cos ^{5 / 2} 2 \theta+C$. Any help would be appreciated.
$$
\begin{aligned}
&\int \sin ^3 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\int \sin ^2 2 \theta \sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\text { Let } u=\sin ^2 2 \theta, d v=\sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&d u=\sin 4 \theta d \theta, \quad v=\int \sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\text { Let } x=\cos 2 \theta \\
&d x=-2 \sin 2 \theta d \theta \\
&\therefore d \theta=\frac{d x}{-2 \sin 2 \theta} \\
&\therefore v=-\frac{1}{2} \int \sqrt{x} d x \\
&=-\frac{1}{2}\left[\frac{2}{3} u^{3 / 2}\right]=-\frac{1}{3} \cos ^{3 / 2} 2 \theta \\
&\therefore \sin ^2 2 \theta\left(-\frac{1}{3} \cos ^{3 / 2} 2 \theta\right)-\int\left(-\frac{1}{3} \cos ^{3 / 2} 2 \theta\right) \cdot \sin 4 \theta d \theta \\
&
\end{aligned}
$$
 A: There are a couple of mistakes in your solutions:

*

*In your first method you write $(1-u^2) u^{\frac12} =u ^{\frac12} - u ^{\color{red}{\frac32}}$, but the second exponent should be $2 + \frac{1}{2} = \frac{4}{2}+ \frac{1}{2} = \color{red}{\frac{5}{2}}$. With this correction, the antiderivative you get is
$$
-\frac{1}{3}\cos^{\frac32}(2\theta) + \frac{1}{7}\cos^{\frac72}(2\theta)+C
$$

*On your second solution you write $u = \sin^2(2\theta)$ implies $\mathrm{d}u = \sin(4\theta)$, but the derivative of $\sin^2(2\theta)$ is $\color{green}{2}\sin(4\theta)$ (you probably forgot to apply the chain rule on the inner-most $2\theta$).


Since $\sin(2x) = 2\sin(x) \cos(x)$ we get
$$
\int \cos^{\frac32}(2\theta) \sin(4\theta) \, \mathrm{d}\theta = 2\int \cos^{\frac52}(2\theta) \sin(2\theta)\, \mathrm{d} \theta \overset{\color{blue}{u = \cos(2\theta)}}{=} - \int u^{\frac52}\, \mathrm{d}u  = - \frac27 \cos^{\frac72}(2\theta) \tag{1}
$$
Then, from your last (corrected) equation, we get
\begin{align*}
\int\sin^3(2\theta)\sqrt{\cos(2\theta)}\, \mathrm{d} \theta  & \overset{\text{IBP}}{=} -\frac{\color{purple}{\sin^2(2\theta)}\cos^{\frac32}(2\theta)}{3} + \frac{\color{green}{2}}{3} \int \cos^{\frac32}(2\theta) \sin(4\theta) \, \mathrm{d}\theta \\
& \overset{(1)}{=} -\frac{\color{purple}{\left(1-\cos^2(2\theta)\right)}\cos^{\frac32}(2\theta)}{3}  - \frac{4}{21} \cos^{\frac72}(2\theta) +C\\
& = -\frac{1}{3}\cos^{\frac32}(2\theta) + \frac{1}{7}\cos^{\frac72}(2\theta)+C
\end{align*}
as desired.
