\(\rm \int_{-\pi}^\pi \frac{2y(1+\sin y)}{1+\cos^2y}dy\) এর মান হল:

ধারণা:

\(\int_a^bf(x) \ dx= \int_a^bf(a+b-x) \ dx= \)

\(\int_{-a}^af(x) \ dx=\left\{\begin{matrix} 0 & if \ f(x) \ is \ odd \\ 2\int_0^af(x) \ dx & if \ f(x) \ is \ even \\ \end{matrix}\right.\)

গণনা:

প্রদত্ত, I = \(\rm \int_{-π}^π \frac{2y(1+\sin y)}{1+\cos^2y}dy\)

= \(\displaystyle \int_{-π}^π \frac{2 y}{1+\cos ^2 y} d y+\int_{-π}^π \frac{2 y \sin y}{1+\cos ^2 y} d y \\\ \rm \ \ \ \ \ \ \\ (Odd) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Even)\)

= \(0+2.2 \displaystyle \int_0^π {y}\left(\frac{\sin y}{1+\cos ^2 y}\right) d y\)

= \(4 \displaystyle \int_0^π \frac{\mathrm{y} \sin \mathrm{y}}{1+\cos ^2 {y}} {dy}\)

= \(4 \displaystyle \int_0^π \frac{(π-y) \sin y}{1+\cos ^2 {y}} {dy}\)

∴ 2I = \(4 \displaystyle \int_0^π \frac{π \sin {y}}{1+\cos ^2 {y}} {dy}\)

⇒ I = \(2 π \int_0^π \frac{\sin {y}}{1+\cos ^2 {y}} {dy}\)

= \(2 π\left(-\tan ^{-1}\left(\cos _y\right)\right)_0^π\)

= \(-2 π\left[\left(-\frac{π}{4}\right)-\left(\frac{π}{4}\right)\right]\)

= \(-2 π\left[-\frac{2 π}{4}\right]=π^2\)

∴ ইন্টিগ্রালের মান হল π².

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