Integrate by parts twice to show that $\int e^{2x}\sin x\,dx=\tfrac{1}{5}e^{2x}(2\sin x-\cos x)+c.$ The diagram below shows the shaded region $R$ that is bounded by part of the curve $y= e^{x-\pi/2}\sqrt{\sin x}$, the line $4x=\pi(y+1)$ and the $x$-axis. The line $4x=\pi(y+1)$ cuts the curve and the $x$-axis at $\left(\dfrac{\pi}{2},1\right)$ and $\left(\dfrac{\pi}{4},0\right)$ respectively. Show that the volume generated when region $R$ is rotated $2\pi$ radians about the $x$-axis is given by $A\pi(B+e^{-\pi})+C\pi r^2$, where $A$, $B$ and $C$ are constants to be determined.

1. Set up and apply the first integration by parts.

Let $I = \int e^{2x}\sin x\,dx$.
Using the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$, we choose $u = e^{2x}$ and $\frac{dv}{dx} = \sin x$.
Then $\frac{du}{dx} = 2e^{2x}$ and $v = -\cos x$.
$I = -e^{2x}\cos x - \int (-\cos x)(2e^{2x})\,dx$
$I = -e^{2x}\cos x + 2\int e^{2x}\cos x\,dx$

2. Apply integration by parts a second time on the new integral.

For the remaining integral $\int e^{2x}\cos x\,dx$, we apply integration by parts again, choosing $u = e^{2x}$ and $\frac{dv}{dx} = \cos x$.
Then $\frac{du}{dx} = 2e^{2x}$ and $v = \sin x$.
$\int e^{2x}\cos x\,dx = e^{2x}\sin x - \int (\sin x)(2e^{2x})\,dx$
$\int e^{2x}\cos x\,dx = e^{2x}\sin x - 2\int e^{2x}\sin x\,dx$

3. Substitute the result of the second integration back to form an equation in terms of the original integral.

Substitute this result back into the expression for $I$:
$I = -e^{2x}\cos x + 2\left(e^{2x}\sin x - 2\int e^{2x}\sin x\,dx\right)$
Recognising that the original integral has reappeared on the right-hand side, we replace $\int e^{2x}\sin x\,dx$ with $I$:
$I = -e^{2x}\cos x + 2e^{2x}\sin x - 4I$

4. Solve algebraically for the original integral to obtain the final expression.

Bring all terms involving $I$ to the left-hand side:
$I + 4I = 2e^{2x}\sin x - e^{2x}\cos x$
$5I = e^{2x}(2\sin x - \cos x)$
Divide by 5 and add an arbitrary constant of integration $c$:
$I = \frac{1}{5}e^{2x}(2\sin x - \cos x) + c$
$\int e^{2x}\sin x\,dx = \tfrac{1}{5}e^{2x}(2\sin x - \cos x) + c$ (shown)

1. Set up the volume as (solid under curve) − (cone under line)

The curve $y=e^{x-\pi/2}\sqrt{\sin x}$ meets the $x$-axis at the origin (where $\sin x=0$) and reaches $(\tfrac{\pi}{2},1)$. The region $R$ is bounded below by the $x$-axis from $x=0$ to $x=\tfrac{\pi}{4}$, on the right by the line from $(\tfrac{\pi}{4},0)$ to $(\tfrac{\pi}{2},1)$, and above by the curve. Rotating about the $x$-axis:
$V=\pi\int_{0}^{\pi/2}y_{\text{curve}}^{2}\,dx-\Big(\text{cone from the line between }x=\tfrac{\pi}{4}\text{ and }x=\tfrac{\pi}{2}\Big).$
Here $y_{\text{curve}}^{2}=e^{2(x-\pi/2)}\sin x=e^{2x-\pi}\sin x$.

2. Evaluate the curve volume using part (a)

$\pi\int_{0}^{\pi/2}e^{2x-\pi}\sin x\,dx=\pi e^{-\pi}\int_{0}^{\pi/2}e^{2x}\sin x\,dx.$
Using part (a), $\int e^{2x}\sin x\,dx=\tfrac{1}{5}e^{2x}(2\sin x-\cos x)$:
$\Big[\tfrac{1}{5}e^{2x}(2\sin x-\cos x)\Big]_{0}^{\pi/2}=\tfrac{1}{5}e^{\pi}(2)-\tfrac{1}{5}e^{0}(-1)=\tfrac{1}{5}(2e^{\pi}+1).$

3. Simplify the curve volume to the required form

$\pi e^{-\pi}\cdot\tfrac{1}{5}(2e^{\pi}+1)=\tfrac{1}{5}\pi\left(2+e^{-\pi}\right).$
This is the $A\pi(B+e^{-\pi})$ term with $A=\tfrac{1}{5}$, $B=2$.

4. Compute the cone volume and combine

The line from $(\tfrac{\pi}{4},0)$ to $(\tfrac{\pi}{2},1)$ generates a cone of radius $r=1$ and height $h=\tfrac{\pi}{2}-\tfrac{\pi}{4}=\tfrac{\pi}{4}$:
$V_{\text{cone}}=\tfrac{1}{3}\pi r^{2}h=\tfrac{1}{3}\pi(1)^{2}\cdot\tfrac{\pi}{4}=\tfrac{\pi^{2}}{12}.$
Therefore
$V=\tfrac{1}{5}\pi\left(2+e^{-\pi}\right)-\tfrac{1}{3}\pi r^{2}h=\tfrac{1}{5}\pi\left(2+e^{-\pi}\right)-\tfrac{\pi^{2}}{12},$
so $A=\tfrac{1}{5}$, $B=2$, $C=-\tfrac{1}{3}$.