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Question Number 50414 by Abdo msup. last updated on 16/Dec/18

c a l c u l a t e \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

c t a n x = \frac{1}{t a n x}

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

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

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

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

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

n o w I_{1} = \int \frac{t a n x}{(1 + {t a n}^{2} x) ({t a n}^{2} x + \frac{1}{{t a n}^{2} x})} d x

t = t a n x d t = {s e c}^{2} x d x

\frac{d t}{1 + t^{2}} = d x

\int \frac{t}{(1 + t^{2}) (t^{2} + \frac{1}{t^{2}})} \times \frac{d t}{1 + t^{2}}

\int \frac{t^{3}}{{(1 + t^{2})}^{2} (t^{4} + 1)} d t

k = t^{2} d k = 2 t d t \frac{d k}{2} = t d t

\int \frac{k \times \frac{d k}{2}}{{(1 + k)}^{2} (1 + k^{2})}

\frac{1}{4} \int \frac{2 k d k}{{(1 + k)}^{2} (1 + k^{2})}

\frac{1}{4} \int \frac{{(1 + k)}^{2} - (1 + k^{2})}{{(1 + k)}^{2} (1 + k^{2})} d k

\frac{1}{4} \int \frac{d k}{1 + k^{2}} - \frac{1}{4} \int \frac{d k}{{(1 + k)}^{2}}

= \frac{1}{4} {t a n}^{- 1} (k) - \frac{1}{4} \times \frac{{(1 + k)}^{- 2 + 1}}{- 2 + 1}

= \frac{1}{4} {t a n}^{- 1} (k) + \frac{1}{4 (1 + k)}

= \frac{1}{4} {t a n}^{- 1} (t^{2}) + \frac{1}{4 (1 + t^{2})}

= \frac{1}{4} {t a n}^{- 1} ({t a n}^{2} x) + \frac{1}{4 (1 + {t a n}^{2} x)}

2 I = \frac{π}{2} \times ∣ \frac{1}{4} {t a n}^{- 1} ({t a n}^{2} x) + \frac{1}{4 (1 + {t a n}^{2} x)} ∣_{0}^{\frac{π}{2}}

2 I = \frac{π}{8} [{({t a n}^{- 1} ({t a n}^{2} \frac{π}{2}) + \frac{1}{1 + {t a n}^{2} \frac{π}{2}})} - {({t a n}^{- 1} (0) + \frac{1}{1 + 0})}]

I = \frac{π}{16} [({t a n}^{- 1} (\infty) + \frac{1}{1 + \infty}) - (1)]

I = \frac{π}{16} (\frac{π}{2} + 0 - 1)

I = \frac{π^{2}}{32} - \frac{π}{16}

p l s c h e c k \dots