Calculus-I-P-0-pi-2-xcos-x-1-e-sin-x-dx-0-pi-2-xsin-x-1-e-cos-x-dx-

Question Number 144527 by mnjuly1970 last updated on 26/Jun/21

\dots . . Calculus (I) \dots . .

P := \frac{\int_{0}^{\frac{π}{2}} (x c o s (x) + 1) e^{s i n (x)} d x}{\int_{0}^{\frac{π}{2}} (x s i n (x) - 1) e^{c o s (x)} d x} = ?

Answered by Kamel last updated on 26/Jun/21

= \frac{{[{x e}^{s i n (x)}]}_{0}^{\frac{π}{2}}}{- {[{x e}^{c o s (x)}]}_{0}^{\frac{π}{2}}} = - e

Answered by Dwaipayan Shikari last updated on 26/Jun/21

\frac{d}{d x} ({x e}^{f (x)}) = {x e}^{f (x)} f^{'} (x) + e^{f (x)}

{x e}^{f (x)} = \int {x e}^{f (x)} f^{'} (x) + e^{f (x)} d x

H e r e f (x) = s i n x g (x) = c o s x

P = - \frac{{[{x e}^{s i n x}]}_{0}^{\frac{π}{2}}}{{[{x e}^{c o s x}]}_{0}^{\frac{π}{2}}} = - \frac{\frac{π e}{2}}{\frac{π}{2}} = - e