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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
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Answer

Consider the integral
$\text{I}=\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
Divide both the numerator and the denominator by $\cos^4\text{x}$, we have,
$\text{I}=\int\frac{\frac{1}{\cos^4\text{x}}}{\frac{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}{\cos^4\text{x}}}\ \text{dx}$
$=\int\frac{\sec^4\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
$=\int\frac{\sec^2\text{x}\times\sec^2\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
$=\int\frac{(\tan^2\text{x}+1)\text{x}\sec^2\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
Substituting $\tan\text{x}=\text{t};\ \sec^2\text{x dx}=\text{dt}$
Thus
$\text{I}=\int\frac{(1+\text{t})^2}{\text{t}^4+\text{t}^2+1}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}^2+\frac{1}{\text{t}^2}+1\Big)}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}^2+\frac{1}{\text{t}^2}-2+2+1\Big)}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}-\frac{1}{\text{t}}\Big)^2+3}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}-\frac{1}{\text{t}}\Big)^2+3}$
Substituting $\text{z}=\text{t}=-\frac{1}{\text{t}};\text{dz}=\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}$
$\text{I}=\int\frac{\text{dz}}{\text{z}^2+3}$
$\Rightarrow\text{I}=\int\frac{\text{dz}}{\text{z}^2+(\sqrt{3})^2}$
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{\text{t}}}{\sqrt{3}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\bigg(\frac{\tan\text{x}-\frac{1}{\tan\text{x}}}{\sqrt{3}}\bigg)+\text{C}$
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\tan\text{x}-\cot\text{x}}{\sqrt{3}}\Big)+\text{C}$

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