Question

Evaluate \[ \int_{0}^{\pi}\frac{x}{\sin x}\left(3\cos^2x+2\sin x+\sin^3x-3\right)\,dx \] is equal to:

• A. \(\dfrac{\pi(5\pi-12)}{4}\)
• B. \(\dfrac{\pi}{2}\)
• C. \(\dfrac{\pi}{2}(5\pi-6)\)
• D. \(\dfrac{\pi(5\pi-12)}{6}\)

Hint:

For integrals of the form \[ \int_{0}^{a}x f(x)\,dx \] where \(f(a-x)=f(x)\), use \[ \int_{0}^{a}x f(x)\,dx=\frac{a}{2}\int_{0}^{a}f(x)\,dx. \]

Answer 1

The Correct Option is A

Step 1: Simplify $ 3\cos^2x-3=-3\sin^2x $.
So $ 3\cos^2x+2\sin x+\sin^3x-3=\sin x(-3\sin x+2+\sin^2x)=\sin x(\sin x-1)(\sin x-2) $.
Step 2: Simplify the full integrand.
\[ \frac{x}{\sin x}\cdot\sin x(\sin x-1)(\sin x-2)=x(\sin^2x-3\sin x+2) \] \[ I=\int_0^\pi x\sin^2x\,dx-3\int_0^\pi x\sin x\,dx+2\int_0^\pi x\,dx=I_1-3I_2+2I_3 \]
Step 3: Evaluate $ I_3=\int_0^\pi x\,dx=\frac{\pi^2}{2} $.
$ I_3=[\frac{x^2}{2}]_0^\pi=\frac{\pi^2}{2} $.
Step 4: Evaluate $ I_2=\int_0^\pi x\sin x\,dx=\pi $ by parts.
$ I_2=[-x\cos x]_0^\pi+[\sin x]_0^\pi=\pi+0=\pi $.
Step 5: Evaluate $ I_1=\int_0^\pi x\sin^2x\,dx=\frac{\pi^2}{4} $ by King property.
$ 2I_1=\pi\int_0^\pi\sin^2x\,dx=\pi\cdot\frac{\pi}{2} \implies I_1=\frac{\pi^2}{4} $.
Step 6: Combine.
\[ I=\frac{\pi^2}{4}-3\pi+\pi^2=\frac{5\pi^2}{4}-3\pi=\frac{\pi(5\pi-12)}{4} \] \[ \boxed{\dfrac{\pi(5\pi-12)}{4}} \]