calculate ∫_0 ^(π/2) ((x sinx cosx)/(tan^2 x +cotan^2 x))dx ctanx =(1/(tanx))


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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
	$I=∫_0 ^(π/2) ((((π/2)−x)cosxsinx)/(cot^2 x+tan^2 x))dx 2I=∫_0 ^(π/2) ((((π/2)−x)cosxsinx+xsinxcosx)/(cot^2 x+tan^2 x))dx 2I=∫_0 ^(π/2) (((π/2)sinxcosx)/(tan^2 x+cot^2 x))dx 2I=(π/2)∫_0 ^(π/2) ((tanxcos^2 x)/(tan^2 x+(1/(tan^2 x))))dx now I_1 =∫((tanx)/((1+tan^2 x)(tan^2 x+(1/(tan^2 x)))))dx t=tanx dt=sec^2 x dx (dt/(1+t^2 ))=dx ∫(t/((1+t^2 )(t^2 +(1/t^2 ))))×(dt/(1+t^2 )) ∫(t^3 /((1+t^2 )^2 (t^4 +1)))dt k=t^2 dk=2tdt (dk/2)=tdt ∫((k×(dk/2))/((1+k)^2 (1+k^2 ))) (1/4)∫((2kdk)/((1+k)^2 (1+k^2 ))) (1/4)∫(((1+k)^2 −(1+k^2 ))/((1+k)^2 (1+k^2 )))dk (1/4)∫(dk/(1+k^2 ))−(1/4)∫(dk/((1+k)^2 )) =(1/4)tan^(−1) (k)−(1/4)×(((1+k)^(−2+1) )/(−2+1)) =(1/4)tan^(−1) (k)+(1/(4(1+k))) =(1/4)tan^(−1) (t^2 )+(1/(4(1+t^2 ))) =(1/4)tan^(−1) (tan^2 x)+(1/(4(1+tan^2 x))) 2I=(π/2)×∣(1/4)tan^(−1) (tan^2 x)+(1/(4(1+tan^2 x)))∣_0 ^(π/2) 2I=(π/8)[{(tan^(−1) (tan^2 (π/2))+(1/(1+tan^2 (π/2))))}−{(tan^(−1) (0)+(1/(1+0)))}] I=(π/(16))[(tan^(−1) (∞)+(1/(1+∞)))−(1)] I=(π/(16))((π/2)+0−1) I=(π^2 /(32))−(π/(16)) pls check...$
	$I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) c o s x s i n x}{{c o t}^{2} x + {t a n}^{2} x} d x$ $2 I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) c o s x s i n x + x s i n x c o s x}{{c o t}^{2} x + {t a n}^{2} x} d x$ $2 I = \int_{0}^{\frac{π}{2}} \frac{\frac{π}{2} s i n x c o s x}{{t a n}^{2} x + {c o t}^{2} x} d x$ $2 I = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{{t a n x c o s}^{2} x}{{t a n}^{2} x + \frac{1}{{t a n}^{2} x}} d x$ $n o w I_{1} = \int \frac{t a n x}{(1 + {t a n}^{2} x) ({t a n}^{2} x + \frac{1}{{t a n}^{2} x})} d x$ $t = t a n x d t = {s e c}^{2} x d x$ $\frac{d t}{1 + t^{2}} = d x$ $\int \frac{t}{(1 + t^{2}) (t^{2} + \frac{1}{t^{2}})} \times \frac{d t}{1 + t^{2}}$ $\int \frac{t^{3}}{{(1 + t^{2})}^{2} (t^{4} + 1)} d t$ $k = t^{2} d k = 2 t d t \frac{d k}{2} = t d t$ $\int \frac{k \times \frac{d k}{2}}{{(1 + k)}^{2} (1 + k^{2})}$ $\frac{1}{4} \int \frac{2 k d k}{{(1 + k)}^{2} (1 + k^{2})}$ $\frac{1}{4} \int \frac{{(1 + k)}^{2} - (1 + k^{2})}{{(1 + k)}^{2} (1 + k^{2})} d k$ $\frac{1}{4} \int \frac{d k}{1 + k^{2}} - \frac{1}{4} \int \frac{d k}{{(1 + k)}^{2}}$ $= \frac{1}{4} {t a n}^{- 1} (k) - \frac{1}{4} \times \frac{{(1 + k)}^{- 2 + 1}}{- 2 + 1}$ $= \frac{1}{4} {t a n}^{- 1} (k) + \frac{1}{4 (1 + k)}$ $= \frac{1}{4} {t a n}^{- 1} (t^{2}) + \frac{1}{4 (1 + t^{2})}$ $= \frac{1}{4} {t a n}^{- 1} ({t a n}^{2} x) + \frac{1}{4 (1 + {t a n}^{2} x)}$ $2 I = \frac{π}{2} \times ∣ \frac{1}{4} {t a n}^{- 1} ({t a n}^{2} x) + \frac{1}{4 (1 + {t a n}^{2} x)} ∣_{0}^{\frac{π}{2}}$ $2 I = \frac{π}{8} [{({t a n}^{- 1} ({t a n}^{2} \frac{π}{2}) + \frac{1}{1 + {t a n}^{2} \frac{π}{2}})} - {({t a n}^{- 1} (0) + \frac{1}{1 + 0})}]$ $I = \frac{π}{16} [({t a n}^{- 1} (\infty) + \frac{1}{1 + \infty}) - (1)]$ $I = \frac{π}{16} (\frac{π}{2} + 0 - 1)$ $I = \frac{π^{2}}{32} - \frac{π}{16}$ $p l s c h e c k . . .$
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