Evaluate: 1 ^ 4 x + ^ 4 x dx - Vidyadip

Question

Evaluate:
$\int\frac{1}{\cos^{4}\text{x} +\sin^{4}\text{x}}\text{dx}$

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Answer

$\text{I} = \int\frac{1}{\cos^{4}\text{x} + \sin^{4}\text{x}}\text{dx}$
dividing numerator and denominator by $\cos^4 x$
$ =\int\frac{\sec^{4}\text{x}}{1 + \tan^{4}\text{x}}\text{dx} = \int\frac{(1 + \tan^{2}\text{x})\sec^{2}\text{x}}{1 + \tan^{4}\text{x}}\text{dx}$
Putting $\tan x = t$
$\therefore\sec^{2}\text{x}\text{ dx} =\text{dt}$
$ = \int\frac{(\text{t}^{2} + 1)\text{dt}}{\text{t}^{4} + 1 } = \int\frac{1 +\frac{1}{\text{t}^{2}}}{\text{t}^{2} + \frac{1}{\text{t}^{2}}}\text{dt}$ $\left\{\text{ dividing by t}^{2}\right\}$
$ = \int\frac{\text{dz}}{\text{z}^{2} + (\sqrt{2})^{2}}\text{ where }\text{t} - \frac{1}{\text{t}} =\text{z}$
$ = \frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\text{z}}{\sqrt{2}}\bigg) +\text{c}$
$ =\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\text{t}^{2} - 1 }{\sqrt{2}\text{t}}\bigg) + \text{c} = \frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\tan^{2}\text{x} - 1}{\sqrt{2}\tan\text{x}}\bigg) +\text{c}.$

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