First slide

Methods of integration

Question

If $\int \frac{d x}{\sin^{4} x + \cos^{4} x} = \frac{1}{\sqrt{2}} \tan^{- 1} f (x) + C$ then

Easy

A

$f (x) = \tan x - \cot x$
B

$f (π / 4) = 0$
C

$f (x)$ is continuous on $R$
D

$f (x) = \frac{1}{2} (\tan x - \cot x)$

Solution

$Dividing the numerator and denominator by$
$\cos^{4} x, the given integral can be written as$
$\begin{array}{l} \int \frac{\sec^{2} x (1 + \tan^{2} x)}{\tan^{4} x + 1} d x = \int \frac{1 + t^{2}}{1 + t^{4}} d t (t = \tan x) \\ = \int \frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2}} d t = \int \frac{1 + 1 / t^{2}}{{(t - \frac{1}{t})}^{2} + 2} d t \\ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (t - \frac{1}{t}) + C \\ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{\tan x - \cot x}{\sqrt{2}} + C \end{array}$
Thus $f (x) = \frac{\tan x - \cot x}{\sqrt{2}} so f (\frac{π}{4}) = 0$

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