0-pi-x-tan-x-sec-x-cos-x-dx-

Question Number 52947 by gunawan last updated on 15/Jan/19

\underset{0}{\int^{π}} \frac{x \tan x}{\sec x + \cos x} d x =

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

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

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

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

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

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

- 2 I = π \times ∣ {t a n}^{- 1} (c o s x) ∣_{0}^{π}

I = (\frac{- π}{2}) \times {{t a n}^{- 1} (- 1) - {t a n}^{- 1} (1)}

= \frac{(- π)}{2} \times {- 2 {t a n}^{- 1} (1)}

= \frac{π}{2} \times 2 \times \frac{π}{4} = \frac{π^{2}}{4} p l s c h e c k \dots

Commented by gunawan last updated on 16/Jan/19

Good bless you Sir

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