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### Answers (1)

Answer:

$-e^{x} \cot x+C$

Given:

$\int e^{x}\left(1-\cot x+\cot ^{2} x\right) d x$

Hint:

Using integration by parts

Explanation:

Let $I=\int e^{x}\left(1-\cot x+\cot ^{2} x\right) d x$

\begin{aligned} &=\int e^{x}\left[\left(1+\cot ^{2} x\right)-\cot x\right] d x \\ &=\int e^{x}\left(\operatorname{cosec}^{2} x-\cot x\right) d x \end{aligned}                                                $\left[\because 1+\cot ^{2} \theta=\cos e c^{2} \theta\right]$

\begin{aligned} &=-\int e^{x}\left(\cot x-\cos e c^{2} x\right) d x \\ =&-\int e^{x} \cot x d x+\int e^{x} \operatorname{cosec}^{2} x d x \\ =&-\left[\cot x \cdot e^{x}-\int\left(-\cos e c^{2} x\right) e^{x} d x\right]+\int e^{x} \cos e c^{2} x d x \ldots .\left\{\int u \cdot v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\} \\ =&-\cot x \cdot e^{x}-\int e^{x} \cos e c^{2} x d x+\int e^{x} \cos e c^{2} x d x \\ =&-\cot x e^{x}+C \quad \text { [Using integration by parts ] } \end{aligned}

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