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

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

Answers (1)

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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