Question

Evaluate the ∫2sec² β bβ/tanβ

Answer 1

Evaluate the $\int \frac{2\sec^2{\beta}d\beta}{\tan{\beta}}$ .

Solution:

$\int \frac{2\sec^2{\beta}d\beta}{\tan{\beta}}=\int \frac{2d\beta}{\tan{\beta}\cos^2{\beta}}=\int \frac{2d\tan{\beta}}{\tan{\beta}}=$

$2\ln{|\tan{\beta}|}+C=\ln{(\tan^2{\beta})}+C$

Answer:$\int \frac{2\sec^2{\beta}d\beta}{\tan{\beta}}=\ln{(\tan^2{\beta})}+C$ .