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Question
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
Options
- \[\frac{1}{3} \tan^2 x + C\]
- \[\frac{1}{2} \tan^2 x + C\]
- \[\frac{1}{3} \tan^3 x + C\]
none of these
MCQ
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Solution
\[\frac{1}{3} \tan^3 x + C\]
\[\text{Let }I = \int\frac{\sin^2 x dx}{\cos^4 x}\]
\[ = \int\frac{\sin^2 x}{\cos^2 x} \times \frac{1}{\cos^2 x}dx\]
\[ = \int \tan^2 x \cdot \sec^2 x dx\]
\[\text{Let }\tan x = t\]
\[ \Rightarrow \sec^2 x dx = dt\]
\[ \therefore I = \int t^2 \cdot dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\tan^3 x}{3} + C ..........\left( \because t = \tan x \right)\]
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