Evaluate the ∫2sec² β bβ/tanβ
Evaluate the ∫2sec2βdβtanβ\int \frac{2\sec^2{\beta}d\beta}{\tan{\beta}}∫tanβ2sec2βdβ .
Solution:
∫2sec2βdβtanβ=∫2dβtanβcos2β=∫2dtanβtanβ=\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}}=∫tanβ2sec2βdβ=∫tanβcos2β2dβ=∫tanβ2dtanβ=
2ln∣tanβ∣+C=ln(tan2β)+C2\ln{|\tan{\beta}|}+C=\ln{(\tan^2{\beta})}+C2ln∣tanβ∣+C=ln(tan2β)+C
Answer:∫2sec2βdβtanβ=ln(tan2β)+C\int \frac{2\sec^2{\beta}d\beta}{\tan{\beta}}=\ln{(\tan^2{\beta})}+C∫tanβ2sec2βdβ=ln(tan2β)+C .
