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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 21 maths textbook solution

Answers (1)

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Answer:
The correct answer is e^{x} \cot x+c
 

Given:

\int e^{x} \frac{\sin x \cos x-1}{\sin ^{2} x} d x

Solution:

        \begin{aligned} &I=\int e^{x} \frac{\sin x \cos x-1}{\sin ^{2} x} d x \\ &I=\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x \\ &I=\int e^{x} \cot x d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \end{aligned}

Consider  \int e^{x} \operatorname{cosec}^{2} x\; d x

 

            \int e^{x} \operatorname{cosec}^{2} x \; d x=-e^{x} \cot x+\int e^{x} \cot x\; d x

            \begin{aligned} &I=\int e^{x} \cot x \; d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \\ &=\int e^{x} \cot x \; d x+e^{x} \cot x-\int e^{x} \cot x \; d x+c \\ &=e^{x} \cot x+c \end{aligned}

So, the correct answer is e^{x} \cot x+c

 

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