$100 \int_{0}^{\pi} e^{-x / \pi} \frac{(1-\cos 2 x)}{2} d x$
$=50\left\{\int_{0}^{\pi} e^{-x / \pi} d x-\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x\right\}$
$I_{1}=\int_{0}^{\pi} e^{-x / \pi} d x=\left[-\pi e^{-x / \pi}\right]_{0}^{\pi}=\pi\left(1-e^{-1}\right)$
$I_{2}=\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x$
$\left.=-\pi \mathrm{e}^{-x / \pi} \cos 2 x\right]_{0}^{\pi}-\int-\pi e^{-x / \pi}(-2 \sin 2 x) d x$
$=\pi\left(1-e^{-1}\right)-2 \pi \int_{0}^{\pi} e^{-x / \pi} \sin 2 x d x$
$\left.=\pi\left(1-e^{-1}\right)-2 \pi\left\{-\pi e^{-x / \pi} \sin 2 x\right]_{0}^{\pi}-\int_{0}^{\pi}-\pi e^{-x / \pi} 2 \cos 2 x d x\right\}$
$=\pi\left(1-\mathrm{e}^{-1}\right)-4 \pi^{2} I_{2}$
$\Rightarrow I_{2}=\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}$
$\therefore I=50\left\{\pi\left(1-e^{-1}\right)-\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}\right\}$
$=\frac{200\left(1-e^{-1}\right) \pi^{3}}{1+4 \pi^{2}}$
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${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$
કે જ્યાં $2k - 1 > 0$ તો ${I_1}/{I_2}$ મેળવો.
$f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$. તો $\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _c a=$..........