First slide

Methods of integration

Question

$\int \frac{\sin 2 t}{\sin^{4} t + \cos^{4} t} d t is equal to$

Moderate

A

$\tan^{- 1} (\tan^{2} t) + C$
B

$- \tan^{- 1} (\tan^{2} t) + C$
C

$\tan^{- 1} (\cot^{2} t) + C$
D

$\cot^{- 1} (\tan^{2} t) + C$

Solution

$Let I = \int \frac{\sin 2 t}{\sin^{4} t + \cos^{4} t} d t = \int \frac{2 \sin t \cos t}{\cos^{4} t [\tan^{4} t + 1]} d t$
$I = \int \frac{2 (\frac{\sin t}{\cos t}) \sec^{2} t d t}{[{(\tan t)}^{4} + 1]}$
$Put \tan^{2} t = x \Rightarrow 2 \tan t \sec^{2} t d t = d x$
$\begin{array}{l} I = \int \frac{d x}{x^{2} + 1} \\ = \tan^{- 1} (x) + C \\ = \tan^{- 1} (\tan^{2} t) + C \end{array}$

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