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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{\sin^3\text{x}\cos^5\text{x}}\text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{\sin^3\text{x}\cos^5\text{x}}\text{dx}\ ...(\text{i})$
Then, $\text{I}=\int\sin^{-3}\text{x}\cos^{-5}\text{x}\text{ dx}$
Since -3 - 5 = -8, which is given integer. So, we divide both numerator and denominator by $\cos^8\text{x}$
$\therefore\ \text{I}=\int\frac{\frac{1}{\cos^8\text{x}}}{\frac{\sin^3\text{x}\cos^5\text{x}}{\cos^8\text{x}}}\text{ dx}$
$=\int\frac{\sec^8\text{x}}{\tan^3\text{x}}\text{ dx}$
$=\int\frac{(\sec^2\text{x})^3}{\tan^3\text{x}}\sec^2\text{x}\text{ dx}$
$=\int\frac{(1+\tan^2\text{x})^3}{\tan^3\text{x}}\sec^2\text{x}\text{ dx}$
$\text{I}=\int\frac{\big(1+\tan^6\text{x}+3\tan^4\text{x}+3\tan^2\text{x}\big)\sec^2\text{x}}{\tan^3\text{x}}\text{ dx}\ ...(\text{ii})$
Let $\text{t}=\tan\text{x}$ Then
$\text{d}(\tan\text{x})=\text{dt}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\therefore\ \text{I}\int\frac{\big(1+\text{t}^6+3\text{t}^4+3\text{t}^2\big)}{\text{t}^3}\text{ dt}$
$=\int\big(\text{t}^{-3}+\text{t}^3+3\text{t}+3\text{t}^{-1}\big)\text{dt}$
$=-\frac{\text{t}^{-2}}{2}+\frac{\text{t}^{4}}{4}+\frac{3}{2}\text{t}^2+3\log\text{t}+\text{C}$
$=-\frac{1}{2\text{t}^{2}}+\frac{\text{t}^{4}}{4}+\frac{3}{2}\text{t}^2+3\log\text{t}+\text{C}$
$=-\frac{1}{2}\times\frac{1}{\tan^2\text{x}}+\frac{\tan^4\text{x}}{4}+\frac{3}{2}\times\tan^2\text{x}+3\log|\tan\text{x}|+\text{C}$
$\therefore\ \text{I}=\frac{-1}{2\tan^2\text{x}}+3\log|\tan\text{x}|+\frac{3}{2}\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$

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