Sum
Evaluate:
\[\int\frac{2 \cos^2 x - \cos 2x}{\cos^2 x}dx\]
Advertisement Remove all ads
Solution
\[\int\left( \frac{2 \cos^2 x - \cos 2x}{\cos^2 x} \right)dx\]
\[ = \int\left( \frac{2 \cos^2 x - \left( 2 \cos^2 x - 1 \right)}{\cos^2 x} \right)dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ = \int \sec^2\text{ x dx}\]
\[ = \ \text{tan x} + C\]
Concept: Evaluation of Definite Integrals by Substitution
Is there an error in this question or solution?
APPEARS IN
Advertisement Remove all ads