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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int e^x. \bigg[\large\frac{1- \sin x}{1- \cos x }\bigg].$$dx$

$(a)\;-e^x cos(x/2)+c \\(b)\;-e^{x} \cot (x/2)+c \\(c)\;-e^x \sin (x/2)+c \\ (d)\;None$
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1 Answer

=> $e^x. \bigg[\large\frac{1}{2 \sin ^2 x/2} - \large \frac {2 \sin x/2 \cos x/2}{2 \sin ^2 x/2}\bigg]$$dx$
=> $\int \large\frac{e^x}{2}$$ . cosec^2 (x/2) dx - \int e^x. \cot x/2 dx$
=> $\int \large\frac{e^x}{2}$$ . cosec^2 (\large \frac{x}{2})$$ dx - \cot (\frac{x}{2}).e^x- \int cosec ^2 x/2 \times \frac{1}{2} . e^x dx$
$-e^{x} \cot (x/2)+c$
Hence b is the correct answer.
answered Dec 30, 2013 by meena.p
 
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