Question

Find the integral.\\ $\int \sec^2 x \tan x \, dx$\\ $\circ \tan 2x + C$\\ $\circ \sec 2x + C$\\ $\circ \frac{1}{2} \tan x + C$\\ $\circ \frac{1}{2} \tan^2 x + C$

          Find the integral.\\
$\int \sec^2 x \tan x \, dx$\\
$\circ \tan 2x + C$\\
$\circ \sec 2x + C$\\
$\circ \frac{1}{2} \tan x + C$\\
$\circ \frac{1}{2} \tan^2 x + C$

Find the integral.

∫sec^2 x tan x dx

∘tan 2x + C

∘sec 2x + C

∘(1)/(2)tan x + C

∘(1)/(2)tan^2 x + C

Added by Alejandro A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Tim Thornhill

Step 1

Step 1: Start with the given integral ∫ sec^2(x)tan(x)dx. Show more…

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Thanks for your feedback!

Find the integral. âˆ« sec^2(x)tan(x)dx = âˆ« tan(x)sec^2(x)dx = âˆ« sin(x)/cos(x) * 1/cos^2(x) dx = âˆ« sin(x)/cos^3(x) dx = âˆ« (sin(x) * cos(x)^-3) dx = âˆ« (sin(x) * (1/cos(x))^3) dx = âˆ« (sin(x) * (1/cos^3(x))) dx = âˆ« (sin(x) * (sec^3(x))) dx = âˆ« sec^3(x)sin(x) dx = -1/2 * sec(x)tan(x) + 1/2 * âˆ« sec(x)dx = -1/2 * sec(x)tan(x) + 1/2 * ln|sec(x) + tan(x)| + C