Use integration by parts to establish the given integrals

Use the previous four exercises (Exercise #7, Exercise #8, Exercise #9, and Exercise #10) to prove the following:

$\displaystyle{ \int x \sin^2 x \, dx = \frac{1}{4} x^2 -\frac{1}{4} x \sin (2x) - \frac{1}{8} \cos (2x)}$ .
$\displaystyle{ \int x \sin^3 x \, dx = \frac{3}{4} \sin x - \frac{1}{36} \sin (3x) - \frac{3}{4} x \cos x + \frac{1}{12} x \cos (3x)}$ .
$\displaystyle{ \int x^2 \sin^2 x \, dx = \frac{1}{6} x^3 + \left( \frac{1}{8} - \frac{1}{4} x^2 \right) \sin (2x) - \frac{1}{4} x \cos (2x)}$ .

Proof. First, to apply integration by parts we define the following functions
$\begin{align*} u &= x & \implies && du &= dx \\ dv &= \sin^2 x \, dx & \implies && v &= \frac{x}{2} - \frac{\sin (2x)}{4}. \end{align*}$

We used Exercise #7 for the primitive of $\sin^2 x$ . So, we then have

$\begin{align*} \int x \sin^2 x \, dx &= \int u \, dv \\[9pt] &= uv - \int v \, du \\[9pt] &= \frac{1}{2} x^2 - \frac{1}{4} x \sin (2x) - \int \left( \frac{x}{2} - \frac{\sin (2x)}{4} \right) \, dx \\[9pt] &= \frac{1}{2} x^2 - \frac{1}{4} x \sin (2x) - \frac{1}{4} x^2 - \frac{1}{8} \cos (2x) \\[9pt] &= \frac{1}{4} x^2 - \frac{1}{4} x \sin (2x) - \frac{1}{8} \cos (2x). \qquad \blacksquare \end{align*}$
Proof. To apply integration by parts we define the following functions
$\begin{align*} u &= x & \implies && du &= dx \\ dv &= \sin^3 x \, dx & \implies && v &= -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x). \end{align*}$

Where we obtained the primitive of $\sin^3 x$ from Exercise #10. Then applying integration by parts we have

$\begin{align*} \int x \sin^3 x \, dx &= \int u \, dv \\[9pt] &= uv - \int v \, du \\[9pt] &= -\frac{3}{4} x \cos x + \frac{1}{12} x \cos (3x) - \int \left(-\frac{3}{4} \cos x + \frac{1}{12} \cos (3x) \right) \, dx \\[9pt] &= -\frac{3}{4} x \cos x + \frac{1}{12} x \cos (3x) + \frac{3}{4} \sin x - \frac{1}{36} \sin (3x) \\[9pt] &= \frac{3}{4} \sin x - \frac{1}{36} \sin (3x) - \frac{3}{4} x \cos x + \frac{1}{12} x \cos (3x). \qquad \blacksquare \end{align*}$
Proof. To apply integration by parts we define the following functions
$\begin{align*} u &= x & \implies && du &= dx \\ dv &= x \sin^2 x \, dx & \implies && v &= \frac{1}{4} x^2 - \frac{1}{4} x \sin (2x) - \frac{1}{8} \cos (2x). \end{align*}$

Where we derived the primitive for $x \sin^2 x$ in part (a) above. So, using integration by parts we have

$\begin{align*} \int x^2 \sin ^2 x \, dx &= \int u \, dv \\[9pt] &= uv - \int v \, du \\[9pt] &= \frac{1}{4} x^3 - \frac{1}{4} x^2 \sin (2x) - \frac{1}{8} x \cos (2x) \\[9pt] & \qquad \qquad - \int \left( \frac{1}{4} x^2 - \frac{1}{4} x \sin (2x) - \frac{1}{8} \cos (2x) \right) \, dx \\[9pt] &= \frac{1}{4} x^3 - \frac{1}{4}x^2 \sin (2x) - \frac{1}{8} x \cos (2x) - \frac{1}{12}x^3 \\[9pt] & \qquad \qquad +\frac{1}{4} \int x \sin (2x) \, dx + \frac{1}{16} \sin (2x). \end{align*}$

Then, we need to use integration by parts again to evaluate $\int x \sin (2x) \, dx$ . Let

$\begin{align*} u &= x & \implies && du &= dx \\ dv &= \sin (2x) \, dx & \implies && v &= -\frac{1}{2} \cos (2x). \end{align*}$

Integrating by parts

$\begin{align*} \int x \sin (2x) \, dx &= \int u \, dv \\[9pt] &= uv - \int v \, du \\[9pt] &= -\frac{1}{2} x \cos (2x) + \frac{1}{2} \int \cos (2x) \, dx \\[9pt] &= -\frac{1}{2} x \cos (2x) + \frac{1}{4} \sin (2x). \end{align*}$

Now, plugging this back in where we left off in evaluating $\int x^2 \sin^2 x \, dx$ we have

$\begin{align*} \int x^2 \sin^2 x \, dx &= \frac{1}{4} x^3 - \frac{1}{4} x^2 \sin (2x) - \frac{1}{8} x \cos (2x) - \frac{1}{12}x^3 \\[9pt] & \qquad \qquad +\frac{1}{4} \int x \sin (2x) \, dx + \frac{1}{16} \sin (2x) \\[9pt] &= \frac{1}{4} x^3 - \frac{1}{4} x^2 \sin (2x) - \frac{1}{8} x \cos (2x) - \frac{1}{12} x^3\\[9pt] & \qquad \qquad +\frac{1}{4} \left( -\frac{1}{2} x \cos (2x) + \frac{1}{4} \sin (2x) \right) + \frac{1}{16} \sin (2x)\\[9pt] &= \frac{1}{4} x^3 - \frac{1}{4} x^2 \sin (2x) - \frac{1}{8} x \cos (2x) - \frac{1}{12} x^3\\[9pt] & \qquad \qquad - \frac{1}{8} x \cos (2x) + \frac{1}{16} \sin (2x) + \frac{1}{16} \sin (2x) \\[9pt] &= \left( \frac{1}{4} - \frac{1}{12} \right) x^3 - \frac{1}{4}x^2 \sin (2x) - \frac{1}{4} x \cos (2x) + \frac{1}{8} \sin (2x) \\ &= \frac{1}{6} x^3 + \left( \frac{1}{8} - \frac{1}{4}x^2 \right) \sin (2x) - \frac{1}{4} x \cos (2x). \qquad \blacksquare \end{align*}$