Question

Consider the integral - tan(0) · ln(3 cos(0)) dė:

Answer 1

$\displaystyle \int -\tan(\theta )\cdot \ln(3\cos(\theta )) ~~ d\theta \\\\[-0.35em] ~\dotfill\\\\ u=\ln(3\cos(\theta ))\implies \cfrac{du}{d\theta }=\cfrac{1}{3\cos(\theta )}\cdot -3\sin(\theta ) \\\\\\ \cfrac{du}{d\theta }=-\tan(\theta )\implies \cfrac{du}{-\tan(\theta )}=d\theta \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \int -\tan(\theta )\cdot u\cdot \cfrac{du}{-\tan(\theta )}\implies \int u\cdot du$