Question

$\int (\sec x)^m (\tan^3 x + \tan x) dx$ is equal to:

• A. $\sec^{m+2} x + C$
• B. $\tan^{m+2} x + C$
• C. $\frac{\sec^{m+2} x}{m+2} + C$
• D. $\frac{\tan^{m+2} x}{m+2} + C$
• E. $\frac{\sec^{m+1} x}{m+1} + C$

Hint:

Whenever an integral contains powers of tangent, check if you can use the $\tan^2 x + 1 = \sec^2 x$ identity to convert everything to secants. If you have a lone $\tan x$ left over after multiplying, it usually completes the $du = \sec x \tan x$ form perfectly.

Answer 1

The Correct Option is C

Concept: The integral involves a product of powers of secant and tangent. We should look for a substitution, usually $u = \sec x$, which has the derivative $du = \sec x \tan x dx$.

Step 1: Simplify the integrand.
Factor out $\tan x$ from the second term: \[ \int (\sec x)^m (\tan^3 x + \tan x) dx = \int (\sec x)^m \tan x (\tan^2 x + 1) dx \] Using the identity $\tan^2 x + 1 = \sec^2 x$: \[ = \int (\sec x)^m \tan x (\sec^2 x) dx = \int (\sec x)^{m+2} \tan x dx \]

Step 2: Rearrange to facilitate substitution.
Pull out one $\sec x$ to form the derivative part $\sec x \tan x$: \[ \int (\sec x)^{m+1} (\sec x \tan x) dx \]

Step 3: Apply substitution and integrate.
Let $u = \sec x$, then $du = \sec x \tan x dx$: \[ \int u^{m+1} du = \frac{u^{m+2}}{m+2} + C \] Substitute back $u = \sec x$: \[ \frac{\sec^{m+2} x}{m+2} + C \]