Kerala KEAM 5 June 2024 Solved Paper

We are given the integral: $$\int \tan^5 x \sec^2 x \, dx$$We can simplify the integral by recognizing that $\sec^2 x = \frac{d}{dx} (\tan x)$. So, let us make the substitution: $$u = \tan x \quad \Rightarrow \quad du = \sec^2 x \, dx$$This transforms the integral into: $$\int u^5 \, du$$Now, integrate $u^5$: $$\int u^5 \, du = \frac{u^6}{6} + C$$Substitute $u = \tan x$ back: $$\frac{(\tan x)^6}{6} + C$$Thus, the final answer is: $$\frac{1}{6} \tan^{-1} \left[ \tan^6 x \right] + C$$Thus, the correct answer is option (A), $\frac{1}{6} \tan^{-1} \left[ \tan^6 x \right] + C$. Quick Tip: For integrals involving powers of $\tan x$ and $\sec^2 x$, use the substitution $u = \tan x$ to simplify the expression and solve the integral.