Question

Q:-integrate this and solve
$∫ \frac{ {sin}^{8} x - {cos}^{8} x}{1 - 2 {sin}^{2}x {cos}^{2} x } dx$...........
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Answer 1

$\green{\bold{\underline{ ☆ UPSC-ASPIRANT ☆} }}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-integrate this and solve

$∫ \frac{ {sin}^{8} x - {cos}^{8} x}{1 - 2 {sin}^{2}x {cos}^{2} x } dx$

$\huge\tt\underline\blue{ANSWER }$

------>>>>Here is your answer<<<<--------

$∫ \frac{ { {(sin}^{4}x) }^{2} - { {(cos}^{4} x)}^{2} }{1 - 2 {sin}^{2}x {cos}^{2}x } dx$

$∫ \frac{ {(sin}^{4} x + {cos}^{4} x )( {sin}^{2}x + {cos}^{2} x) ( {sin}^{2}x - {cos}^{2} x) }{1 - 2 {sin}^{2}x {cos}^{2}x }dx$

$∫ \frac{(1 - 2 {sin}^{2} x {cos}^{2}x)( {sin}^{2} x - {cos}^{2}x) }{(1 - 2 {sin}^{2} x {cos}^{2}x) } dx$

$∫ - ( {sin}^{2} x - {cos}^{2} x)$

$∫ - cos2xdx$

=$- \frac{sin2x}{2} + c$✓

HOPE IT HELPS YOU..

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Answer 2

$\frac{Sin⁸x - Cos⁸x}{1 - 2Sin²x*Cos²x}$

$\frac{(Sin⁴x-Cos⁴x)(Sin⁴x + Cos⁴x)}{1 - 2Sin²x*Cos²x}$

$\frac{(Sin²x+Cos²x)(Sin²x-Cos²x)(Sin⁴x+Cos⁴x)}{1 - 2Sin²x*Cos²x}$

$\frac{(Sin²x - Cos²x)(Sin⁴x + Cos⁴x)}{1 - 2Sin²x*Cos²x}$

$\frac{(Sin²x - Cos²x)[(Sin²x + Cos²x)² - 2Sin²x*Cos²x]}{(1 - 2Sin²x*Cos²x)}$

$\frac{(Sin²x - Cos²x)(1 - 2Sin²x*Cos²x)}{(1 - 2Sin²x*Cos²x)}$

$(Sin²x - Cos²x)$

$Now ,$

$Integration \: of \: (Sin²x - Cos²x)dx$

$Integration \: of \: -(Cos²x - Sin²x)dx$

$Integration \: of \: - Cos2x*dx$

$=> \frac{-Sin2x}{2} + c$