Q37 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

$\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} dx$ is equal to (where $C$ is an arbitrary constant)

A
$\frac{1}{2} sin2x + C$
Correct answer
B
$- \frac{1}{2} sin2x + C$
C
$- \frac{1}{2} sinx + C$
D
$- \sin^{2} x + C$

Solution

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

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

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

$= \int - cos2x dx = - \frac{1}{2} sin2x + C$

Hence, the correct answer is option: 'B'.

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Solution

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$\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} dx$ is equal to (where $C$ is an arbitrary constant)