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Question Number 32708 by abdo imad last updated on 31/Mar/18

l e t g i v e f (x) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + x t a n t)}{t a n t} d t

f i n d a s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{l n (1 + 2 t a n t)}{t a n t} d t .

Commented by abdo imad last updated on 03/Apr/18

w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{\partial}{\partial x} (\frac{l n (1 + x t a n t)}{t a n t}) d t

= \int_{0}^{\frac{π}{2}} \frac{t a n t}{(1 + x t a n t) t a n t} d t = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x t a n t}

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

g i v e f^{,} (x) = \int_{0}^{1} \frac{\frac{1 - u^{2}}{1 + u^{2}}}{\frac{1 - u^{2}}{1 + u^{2}} + x \frac{(2 u)}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

f^{^{'}} (x) = \int_{0}^{1} \frac{2 (1 - u^{2})}{(1 + u^{2}) (1 - u^{2} + 2 x u)} d u

= \int_{0}^{1} \frac{2 (u^{2} - 1)}{(1 + u^{2}) (u^{2} - 2 x u - 1)} d u l e t d e c o m p o s e

F (u) = \frac{2 (u^{2} - 1)}{(1 + u^{2}) (u^{2} - 2 x u - 1)}

u^{2} - 2 x u - 1 \Rightarrow Δ^{^{'}} = x^{2} + 1 \Rightarrow u_{1} = x + \sqrt{1 + x^{2}}

u_{2} = x - \sqrt{1 + x^{2}}

F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1} = \frac{2 (u^{2} - 1)}{(u - u_{1}) (u - u_{2}) (u^{2} + 1)}

a = \frac{2 (u_{1}^{2} - 1)}{(u_{1} - u_{2}) (u_{1}^{2} + 1)} = \frac{2 ({(x + \sqrt{1 + x^{2}})}^{2} - 1)}{2 \sqrt{1 + x^{2} ({(x + \sqrt{1 + x^{2}})}^{2} + 1)}}

b = \frac{2 (u_{2}^{2} - 1)}{(u_{2} - u_{1}) (u_{2}^{2} + 1)} = \frac{2 ({(x - \sqrt{1 + x^{2}})}^{2} - 1)}{- 2 \sqrt{2 + x^{2} ((x - \sqrt{{1 + x^{2})}^{2}} + 1)}}

b e c o n t i n u e d \dots .