MHT CET · Maths · Indefinite Integration
If \(I=\int e^{\sin \theta}\left(\log \sin \theta+\operatorname{cosec}^2 \theta\right) \cos \theta d \theta\), then \(I\) is equal to
- A \(\mathrm{e}^{\sin \theta}\left(\log \sin \theta+\operatorname{cosec}^2 \theta\right)+\mathrm{c}\), (where c is a constant of integration)
- B \(\mathrm{e}^{\sin \theta}(\log \sin \theta+\operatorname{cosec} \theta)+\mathrm{c}\), (where c is a constant of integration)
- C \(\mathrm{e}^{\sin \theta}(\log \sin \theta-\operatorname{cosec} \theta)+\mathrm{c}\), (where c is a constant of integration)
- D \(\mathrm{e}^{\sin \theta}\left(\log \sin \theta-\operatorname{cosec}^2 \theta\right)+\mathrm{c}, \quad(\) where c is a constant of integration)
Answer & Solution
Correct Answer
(C) \(\mathrm{e}^{\sin \theta}(\log \sin \theta-\operatorname{cosec} \theta)+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \therefore \quad & \text { Let } \sin \theta=t \\ \therefore \quad & \cos \theta d \theta=d t \\ \therefore \quad & =\int e^t\left(\log t+\frac{1}{t^2}\right) d t \\ & =\int e^t \log t d t+\int \frac{e^t}{t^2} d t \\ & =\log t e^t-\int \frac{e^t}{t} d t+\left[\frac{e^t}{-t}-\int \frac{e^t}{-t} d t\right]+c \\ & =\log t e^t-\int \frac{e^t}{t} d t-\frac{e^t}{t}+\int \frac{e^t}{t} d t+c \\ & =e^t\left[\log t-\frac{1}{t}\right]+c \\ & =e^{\sin \theta}[\log \sin \theta-\operatorname{cosec} \theta]+c\end{aligned}\)
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