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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 4 maths

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

 

Given: \int e^{x} \cdot\left(\cot x-\operatorname{cosec}^{2} x\right) d x

Solution:

    I=\int e^{x} \cdot\left(\cot x-\operatorname{cosec}^{2} x\right) d x

On using integration by parts, \int u . v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]

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

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

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

 

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