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  • The value of the integral  <img alt="\frac{48}{\pi^{4}} \int_{0}^{\pi}\left(\frac{3 \pi x^{2}}{2}-x^{3}\right) \frac{\sin x}{1+\cos ^{2} x} \mathrm{~d} x" src="https://entrancecorner.oncodecog

The value of the integral  \frac{48}{\pi^{4}} \int_{0}^{\pi}\left(\frac{3 \pi x^{2}}{2}-x^{3}\right) \frac{\sin x}{1+\cos ^{2} x} \mathrm{~d} xis equal to_________.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{ I=\frac{48}{\pi^{4}} \int_{0}^{\pi}\left(\frac{3 \pi x^{2}}{2}-x^{3}\right) \frac{\sin x}{1+\cos ^{2} x} d x} \\

\mathrm{I=\frac{48}{\pi^{4}} \int_{0}^{\pi} x^{2}\left(\frac{3 \pi}{2}-x\right) \frac{\sin x}{1+\cos ^{2} x} d x}     .............(1)

Apply King Property

\mathrm{I=\frac{48}{\pi^{4}} \int_{0}^{\pi}(\pi-x)^{2}\left(\frac{\pi}{2}+x\right) \frac{\sin x}{1+\cos ^{2} x} d x}      .........(2)

Eqn (1) + (2)

\mathrm{I=\frac{12}{\pi^{3}} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x}\left[\pi^{2}+(\pi-2) x(\pi-2x)\right] d x}.....(3)

Apply Kings's Property again

\mathrm{I=\frac{12}{\pi^{3}} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x}\left[\pi^{2}+(\pi-2)(\pi-x)(2 x-\pi)\right] d x}.............(4)

Adding eqn (3) and (4)

\mathrm{I=\frac{6}{\pi^{2}} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x}\left[2\pi+(\pi-2)(\pi-2x)\right] d x}...............(5)

Apply Kings Property again

\mathrm{I=\frac{6}{\pi^{2}} \int_{0}^{\pi} \frac{\sin x}{\left(1+\cos ^{2} x\right)}[2 \pi+(\pi-2)(2 x-\pi)] d x}    ...........(6)

Eqn (5)+(6)

\mathrm{I=\frac{12}{\pi} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x}\\

\mathrm{Let \cos x=t \Rightarrow \sin x d x=-d t}\\

\mathrm{I =\frac{12}{\pi} \int_{1}^{-1} \frac{-d t}{1+t^{2}}=\frac{-12}{\pi}\left[\tan ^{-1} -1-\tan ^{-1} 1\right]=\frac{-12}{\pi} \times \frac{-2 \pi}{4}} \\

\mathrm{=6}

Hence answer is \mathrm{6}

Posted by

vinayak

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