find-the-value-of-0-pi-2-xdx-1-cosx-

Question Number 37071 by math khazana by abdo last updated on 08/Jun/18

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{x d x}{1 + c o s x}

Commented by math khazana by abdo last updated on 10/Jun/18

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int_{0}^{1} \frac{2 a r c t a n (t)}{1 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 4 \int_{0}^{1} \frac{a r c t a n (t)}{1 + t^{2} + 1 - t^{2}} d t

= 2 \int_{0}^{1} a r c t a n (t) d t a n d b y p a r t s

I = 2 {{[t a r c t a n (t)]}_{0}^{1} - \int_{0}^{1} \frac{t}{1 + t^{2}} d t}

= 2 {\frac{π}{4} - \frac{1}{2} {[l n (1 + t^{2})]}_{0}^{1}}

= \frac{π}{2} - l n (2)

★ I = \frac{π}{2} - l n (2) ★

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

\int_{0}^{\frac{Π}{2}} \frac{x}{2 {c o s}^{2} \frac{x}{2}} d x

\frac{1}{2} \int_{0}^{\frac{Π}{2}} \frac{{x s e c}^{2} \frac{x}{2}}{} d x

l e t I_{1} = \int {x s e c}^{2} \frac{x}{2} d x

= x \int {s e c}^{2} \frac{x}{2} d x - \int [\frac{d x}{d x} \int {s e c}^{2} \frac{x}{2} d x] d x

= x \frac{t a n \frac{x}{2}}{\frac{1}{2}} - \int \frac{t a n \frac{x}{2}}{\frac{1}{2}} d x

= x \times 2 t a n \frac{x}{2} - 4 l n ∣ s e c \frac{x}{2} ∣

s o r e q u i r e d a n s i s

\frac{1}{2} {2 x t a n \frac{x}{2} - 4 l n ∣ s e c \frac{x}{2} ∣}_{0}^{\frac{Π}{2}}

= \frac{1}{2} [{2 \times \frac{Π}{2} \times 1 - 4 l n (\sqrt{2})} - {2 \times 0 \times 0 - 4 l n 1}]

= \frac{Π}{2} - 2 \times \frac{1}{2} l n 2

= \frac{Π}{2} - l n 2

Commented by math khazana by abdo last updated on 10/Jun/18

c o r r e c t a n s w e r t h a n k s s i r T a n m a y .

Answered by MJS last updated on 10/Jun/18

\int \frac{x}{1 + \cos x} d x =

[\begin{matrix} \int u^{'} v = u v - \int {u v}^{'} \\ u^{'} = \frac{1}{1 + \cos x} \Rightarrow u = \frac{\sin x}{1 + \cos x} \\ [\begin{matrix} u = \int \frac{d x}{1 + \cos x} = \int \frac{\tan \frac{x}{2}}{\sin x} d x = \\ [t = \frac{x}{2} \to d x = 2 d t] \\ = 2 \int \frac{\tan t}{\sin 2 t} d t = \int \frac{\tan t}{\sin t \cos t} d t = \int \sec^{2} t d t = \\ = \tan t = \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} \end{matrix}] \\ v = x \Rightarrow v^{'} = 1 \end{matrix}]

= \frac{x \sin x}{1 + \cos x} - \int \frac{\sin x}{1 + \cos x} d x =

[t = 1 + \cos x \to d x = - \frac{d t}{\sin x}]

= \frac{x \sin x}{1 + \cos x} + \int \frac{d t}{t} = \frac{x \sin x}{1 + \cos x} + \ln t =

= \frac{x \sin x}{1 + \cos x} + \ln (1 + \cos x) + C

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