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 the expression inside the bracket.
We have \[ 3\cos^2x+2\sin x+\sin^3x-3 \] Using \[ \cos^2x=1-\sin^2x, \] we get \[ 3\cos^2x-3=3(1-\sin^2x)-3 \] \[ =3-3\sin^2x-3 \] \[ =-3\sin^2x \] Therefore, \[ 3\cos^2x+2\sin x+\sin^3x-3 = -3\sin^2x+2\sin x+\sin^3x \] \[ = \sin x(-3\sin x+2+\sin^2x) \]

Step 2: Cancel \(\sin x\).
The integral becomes \[ \int_{0}^{\pi}x(2-3\sin x+\sin^2x)\,dx \] So, \[ I=\int_{0}^{\pi}2x\,dx-3\int_{0}^{\pi}x\sin x\,dx+\int_{0}^{\pi}x\sin^2x\,dx \]

Step 3: Evaluate each integral.
First, \[ \int_{0}^{\pi}2x\,dx = \left[x^2\right]_{0}^{\pi} = \pi^2 \] Now, \[ \int_{0}^{\pi}x\sin x\,dx \] Using integration by parts: \[ \int x\sin x\,dx=-x\cos x+\sin x \] Therefore, \[ \int_{0}^{\pi}x\sin x\,dx = \left[-x\cos x+\sin x\right]_{0}^{\pi} \] \[ = -\pi\cos\pi+\sin\pi \] \[ = \pi \] Also, since \(\sin^2x\) is symmetric about \(\frac{\pi}{2}\), \[ \int_{0}^{\pi}x\sin^2x\,dx = \frac{\pi}{2}\int_{0}^{\pi}\sin^2x\,dx \] Now, \[ \int_{0}^{\pi}\sin^2x\,dx=\frac{\pi}{2} \] Hence, \[ \int_{0}^{\pi}x\sin^2x\,dx = \frac{\pi}{2}\cdot \frac{\pi}{2} = \frac{\pi^2}{4} \]

Step 4: Substitute all values.
\[ I=\pi^2-3(\pi)+\frac{\pi^2}{4} \] \[ I=\frac{5\pi^2}{4}-3\pi \] \[ I=\frac{5\pi^2-12\pi}{4} \] \[ I=\frac{\pi(5\pi-12)}{4} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{\pi(5\pi-12)}{4}} \]