Mathematics

Evaluate the given integral.
$$\int { { e }^{ x }\left( 1-\cot { x } +\cot ^{ 2 }{ x }  \right)  } dx$$


ANSWER

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


SOLUTION
$$I=\displaystyle\int{{e}^{x}\left(1-\cot{x}+{\cot}^{2}{x}\right)dx}$$

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

$$I=\displaystyle\int{{e}^{x}{cosec}^{2}{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}$$

Integrating by parts, we get

$$dv={cosec}^{2}{x}dx\Rightarrow\,v=-\cot{x}$$

$$u={e}^{x}\,\Rightarrow\,du={e}^{x}dx$$

$$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ] $$......by parts formula

$$I=-{e}^{x}\cot{x}-\displaystyle\int{-\cot{x}{e}^{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}+c$$

$$I=-{e}^{x}\cot{x}+\displaystyle\int{\cot{x}{e}^{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}+c$$

$$I=-{e}^{x}\cot{x}+c$$
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Single Correct Medium Published on 17th 09, 2020
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