solve-0-pi-xcosx-1-sin-2-x-dx-

Question Number 118242 by mathdave last updated on 16/Oct/20

s o l v e

\int_{0}^{π} \frac{x \cos x}{(1 + \sin^{2} x)} d x

Answered by TANMAY PANACEA last updated on 16/Oct/20

★ \int \frac{c o s x}{1 + {s i n}^{2} x} d x = \int \frac{d (s i n x)}{1 + {s i n}^{2} x} = {t a n}^{- 1} (s i n x) ★

n o w

\int_{0}^{π} \frac{x c o s x}{1 + {s i n}^{2} x} d x

∣ {x t a n}^{- 1} (s i n x) ∣_{0}^{π} - \int_{0}^{π} 1 \times {t a n}^{- 1} (s i n x) d x

0 - \int_{0}^{π} {t a n}^{- 1} (s i n x) d x

f (x) = {t a n}^{- 1} (s i n x)

f (0) = 0

f (π) = 0

s o {t a n}^{- 1} (s i n x) m a k e a l o o p i n x \in [0, π]

i n b e t w e e n [0, π] {t a n}^{- 1} (s i n x) m u s t h a v e m a x i m u m

v a l u e \dots a t x = \frac{π}{2} {t a n}^{- 1} (s i n \frac{π}{2}) = \frac{π}{4}

\int_{0}^{π} \frac{π}{4} d x > \int_{0}^{π} {t a n}^{- 1} (s i n x) d x > 0

\frac{π^{2}}{4} > \int_{0}^{π} {t a n}^{- 1} (s i n x) d x > 0

(- 1) \frac{π^{2}}{4} < \int_{0}^{π} - {t a n}^{- 1} (s i n x) d x < 0

s o I = \int_{0}^{π} \frac{x c o s x}{(1 + {s i n}^{2} x)} = \int_{0}^{π} - {t a n}^{- 1} (s i n x) d x

\frac{- π^{2}}{4} < I < 0