Question

The integral \( \int \tan^5 x \sec^2 x \, dx \) is equal to:

• A. \( \frac{1}{6} \tan^{-1} \left[ \tan^6 x \right] + C \)
• B. \( \frac{1}{2} \tan^{-1} \left[ \tan^6 x \right] + C \)
• C. \( \frac{1}{4} \tan^{-1} \left[ \tan^4 x \right] + C \)
• D. \( \frac{1}{3} \tan^{-1} \left[ \tan^3 x \right] + C \)
• E. \( \frac{1}{7} \tan^{-1} \left[ \tan^7 x \right] + C \)

Hint:

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.

Answer 1

The Correct Option is A

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), \( \frac{1}{6} \tan^{-1} \left[ \tan^6 x \right] + C \).