Calculate $\int y \sec ^{2} y d y$ .
A) $y \tan y+\ln |\cos y|+C$
B) $y \tan y-\ln |\cos y|+C$
C) $\frac{y^{2}}{2}-\tan y+C$
D) $\frac{y^{2}}{2} \tan y+C$

Question

Quizplus · Accepted Answer

The integral can be solved using integration by parts, where $ u = y $ and $ dv = \sec^2 y \, dy $. This gives $ du = dy $ and $ v = \tan y $. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get $ y \tan y - \int \tan y \, dy $. The integral of $\tan y$ is $-\ln |\cos y|$, leading to the result $ y \tan y + \ln |\cos y| + C $.

Calculate ∫ysec⁡2ydy\int y \sec ^{2} y d y∫ysec2ydy A) ytan⁡y+ln⁡∣cos⁡y∣+Cy \tan y+\ln |\cos y|+Cytany+ln∣cosy∣+C

Calculate $\int y \sec ^{2} y d y$
A) $y \tan y+\ln |\cos y|+C$