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Question Number 44575 by Raj Singh last updated on 01/Oct/18

Commented by $@ty@m last updated on 01/Oct/18

$s i m i l a r t o Q . N o . 41703$
Commented by Raj Singh last updated on 02/Oct/18

$t h i s i s s i n 2 Θ$
Commented by $@ty@m last updated on 02/Oct/18

$i t {d o e s n}^{'} t m a t t e r .$ $\sin 2 θ = 2 \sin θ \cos θ$ $w h e n d i v i d e d b y \cos^{4} θ,$ $N^{r} = 2 \tan θ . \sec^{2} θ$ $. . . . a n d t h e n t h e p r o c e d u r e w i l l$ $b e s i m i l a r .$ $. . . . o u r u l t i m a t e a i m i s t o g e t \tan θ .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Oct/18
	$∫_0 ^(π/4) (((sin^2 θ)/(cos^4 θ))/(1+tan^4 θ))dθ ∫_0 ^(π/4) ((tan^2 θsec^2 θ)/(1+tan^4 θ))dθ t=tanθ dt=sec^2 θ dθ ∫_0 ^1 ((t^2 ×dt)/(1+t^4 )) ∫_0 ^1 (dt/(t^2 +(1/t^2 ))) (1/2)∫_0 ^1 ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt (1/2)∫_0 ^1 ((d(t+(1/t)))/((t+(1/t))^2 −2))+(1/2)∫_0 ^1 ((d(t−(1/t)))/((t−(1/t))^2 +2)) (1/2){(1/(2(√2) ))∣ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))∣_0 ^1 +(1/(√2))∣tan^(−1) (((t−(1/t))/(√2)))∣_0 ^1 } (1/2){(1/(2(√2)))∣ln(((t^2 +1−(√2))/(t^2 +1+(√(2 )))))∣_0 ^1 +(1/(√2))∣tan^(−1) (((t^2 −1)/(√2)))∣_0 ^1 } (1/2){(1/(2(√2)))∣ln(((2−(√2))/(2+(√2))))−ln(((1−(√2))/(1+(√2))))∣+(1/(√2))tan^(−1) (0)−tan^(−1) (((−1)/(√2)))} (1/2){(1/(2(√2)))∣ln((((√2) −1)/((√2) +1)))−ln(((1−(√2))/(1+(√2))))+(1/(√2))(tan^(−1) ((1/(√2)))}$
	$\int_{0}^{\frac{π}{4}} \frac{\frac{{s i n}^{2} θ}{{c o s}^{4} θ}}{1 + {t a n}^{4} θ} d θ$ $\int_{0}^{\frac{π}{4}} \frac{{t a n}^{2} θ {s e c}^{2} θ}{1 + {t a n}^{4} θ} d θ$ $t = t a n θ d t = {s e c}^{2} θ d θ$ $\int_{0}^{1} \frac{t^{2} \times d t}{1 + t^{4}}$ $\int_{0}^{1} \frac{d t}{t^{2} + \frac{1}{t^{2}}}$ $\frac{1}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t$ $\frac{1}{2} \int_{0}^{1} \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - 2} + \frac{1}{2} \int_{0}^{1} \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2}$ $\frac{1}{2} {\frac{1}{2 \sqrt{2}} ∣ l n (\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}) ∣_{0}^{1} + \frac{1}{\sqrt{2}} ∣ {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) ∣_{0}^{1}}$ $\frac{1}{2} {\frac{1}{2 \sqrt{2}} ∣ l n (\frac{t^{2} + 1 - \sqrt{2}}{t^{2} + 1 + \sqrt{2}}) ∣_{0}^{1} + \frac{1}{\sqrt{2}} ∣ {t a n}^{- 1} (\frac{t^{2} - 1}{\sqrt{2}}) ∣_{0}^{1}}$ $\frac{1}{2} {\frac{1}{2 \sqrt{2}} ∣ l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) - l n (\frac{1 - \sqrt{2}}{1 + \sqrt{2}}) ∣ + \frac{1}{\sqrt{2}} {t a n}^{- 1} (0) - {t a n}^{- 1} (\frac{- 1}{\sqrt{2}})}$ $\frac{1}{2} {\frac{1}{2 \sqrt{2}} ∣ l n (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}) - l n (\frac{1 - \sqrt{2}}{1 + \sqrt{2}}) + \frac{1}{\sqrt{2}} ({t a n}^{- 1} (\frac{1}{\sqrt{2}})}$
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