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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate:
$\int\frac{1}{\sin^{4}\text{x} +\sin^{2}\text{x}\cos^{2}\text{x}+\cos^{4}\text{x}}\text{dx}$
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Answer

$\text{I}=\int\frac{1}{\sin^{4}\text{x} +\sin^{2}\text{x}\cos^{2}\text{x}+\cos^{4}\text{x}}\text{dx}$
$ =\int\frac{(\tan^2\text{x}+1)\sec^{2}\text{x}}{\tan^{4}\text{x}+\tan^{2}\text{x}+1}\text{dx} , [$dividing $N \& D$ by $\cos^4 x]$
$ = \int\frac{(\text{t}^{2} + 1)}{\text{t}^{4} +\text{t}^2+ 1 } \text{dx},$ where tan x = t
$ = \int\frac{1+\frac{1}{\text{t}^{2}}}{\text{t}^{2} +\frac{1}{\text{t}^2}+ 1 } \text{dx},[$dividing $N \& D$ by $t^2]$
Putting $\text{t}-\frac{1}{\text{t}}$ = z so that $\Big(\text{t}-\frac{1}{\text{t}^2}\Big)$ $dt = dz$
$\text{and}\ \text{t}^2-\frac{1}{\text{t}^2}=\text{z}^2+2$
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+(\sqrt3)^2}$
$\frac{\text{1}}{\sqrt3}\tan^{-1}\frac{\text{z}}{\sqrt3}+\text{c}=\frac{\text{1}}{\sqrt3}\tan^{-1}\Big(\frac{\text{t}^2-1}{\sqrt3\ \text{t}}\Big)+\text{c}\frac{\text{1}}{\sqrt3}\tan^{-1}\Big(\frac{\tan^2\text{x}-1}{\sqrt3\tan\text{x}}\Big)+\text{c}$

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