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∫ Sin 2 X Cos 4 X D X = - Mathematics

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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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Chapter 19: Indefinite Integrals - MCQ [Page 201]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
MCQ | Q 23 | Page 201

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