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Question Number 30180 by abdo imad last updated on 17/Feb/18

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

Commented by abdo imad last updated on 24/Feb/18

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

= \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{d t}{{t a n}^{2} t + {c o t a n}^{2} t} - \int_{0}^{\frac{π}{2}} \frac{s i n t ? c o s t}{{t a n}^{2} t + {c o t a n}^{2} t} d t \Rightarrow

2 I = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{d t}{{t a n}^{2} t + {c o t a n}^{2} t} d t t h e c h . t a n t = u g i v e

\int_{0}^{\frac{π}{2}} \frac{d t}{{t a n}^{2} t + {c o t a n}^{2} t} = \int_{0}^{\infty} \frac{1}{u^{2} + \frac{1}{u^{2}}} \frac{d u}{1 + u^{2}}

= \int_{0}^{\infty} \frac{u^{2}}{(1 + u^{2}) (1 + u^{4})} d u = \int_{0}^{\infty} \frac{1 + u^{2} - 1}{(1 + u^{2}) (1 + u^{4})} d u

= \int_{0}^{\infty} \frac{d u}{1 + u^{4}} - \int_{0}^{\infty} \frac{d u}{(1 + u^{2}) (1 + u^{4})} t h e c h . u^{4} = x g i v e

\int_{0}^{\infty} \frac{d u}{1 + u^{4}} = \int_{0}^{\infty} \frac{\frac{1}{4} x^{\frac{1}{4} - 1}}{1 + x} d x = \frac{1}{4} \frac{π}{s i n (\frac{π}{4})}

= \frac{π}{4} \frac{1}{\frac{1}{\sqrt{2}}} = \frac{π \sqrt{2}}{4} . l e t d e c o m p o s e F (x) = \frac{1}{(1 + u^{2}) (1 + u^{4})}

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

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

= \frac{a u + b}{1 + u^{2}} + \frac{c u + d}{u^{2} - \sqrt{2} u + 1} + \frac{e u + f}{u^{2} + \sqrt{2} u + 1} \dots . b e c o n t i n u e d \dots