First slide

Methods of integration

Question

$Evaluate \int \frac{1}{\sin^{4} x + \cos^{4} x} d x$

Moderate

A

$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{\tan^{2} x - 1}{\sqrt{2} \tan x}) + C$
B

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan^{2} x - 1}{\tan x}) + C$
C

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan^{2} x - 1}{\sqrt{2} \tan x}) + C$
D

None of these

Solution

$\begin{matrix} I = \int \frac{1}{\sin^{4} x + \cos^{4} x} d x = \int \frac{\frac{1}{\cos^{4} x}}{\frac{\sin^{4} x + \cos^{4} x}{\cos^{4} x}} d x \\ = \int \frac{\sec^{4} x}{\tan^{4} x + 1} d x = \int \frac{\sec^{2} x \sec^{2} x}{\tan^{4} x + 1} d x \\ = \int (\frac{1 + \tan^{2} x}{1 + \tan^{4} x}) \sec^{2} x d x \end{matrix}$
$Putting \tan x = t and \sec^{2} x d x = d t, we get$
$I = \int \frac{1 + t^{2}}{1 + t^{4}} d t D i v i d e b y t^{2} \int \frac{\frac{1}{t^{2}} + 1}{\frac{1}{t^{2}} + t^{2}} d t = T a n^{- 1} (t - \frac{1}{t}) + c \cdot here t - \frac{1}{t} = u, (1 + \frac{1}{t^{2}}) d t = d u \int \frac{d u}{u^{2} + 2}$
$\begin{aligned} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t^{2} - 1}{\sqrt{2} t}) + C \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan^{2} x - 1}{\sqrt{2} \tan x}) + C \end{aligned}$

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