Evaluate $\int x\frac{\tan{x}}{\cos^{2}{x}}\,dx$

Question

What is $\int x \frac {\tan{x}} {\cos^{2}{x}}dx$? I managed to get to $\frac {x\sec^2x}2+\int \frac {\sec^2x}2-\frac12+c$ but I'm not sure if I can absorb the $-\frac12$ into the constant at this point. Guidance would be appreciated.

Answer 1

$$ \int x \frac{\tan(x)}{\cos^2(x)} \; dx $$

$$ \int x \tan(x)\sec^2(x) \; dx $$

Do integration by parts with $u=x$ and $dv = \sec^2(x) \tan(x) \; dx$.

Then $du = dx$ and $v = \frac{1}{2} \sec^2(x)$ or $\frac{1}{2} \tan^2(x)$ depending on how you choose to take the antiderivative. It seems a little easier to choose the former.

Apply the formula for integration by parts to get

$$ \frac{1}{2} x \sec^2(x) - \int \frac{1}{2} \sec^2(x) \; dx$$

$$ \frac{1}{2} x \sec^2(x) - \frac{1}{2} \tan(x) + C$$