Question

integration of e^x(1-sinx)/(1-cosx)

Answer 1

Step-by-step explanation:

Given:

${e}^{x} . \frac{1 - sinx}{1 - cosx}$

To find: Integration of the function.

Solution:

We know that

$\int {e}^{x} (f(x) + f'(x))dx = {e}^{x} f(x) + C$

So,try to convert this function in the form stated above.

$\int{e}^{x} . \frac{1 - sinx}{1 - cosx}dx \\ \\ we \: know \: that \\ 1 - cosx = 2 {sin}^{2} \frac{x}{2} \\ \\ sinx = 2sin \frac{x}{2} cos \frac{x}{2}$

Split the term and apply identities

$\int{e}^{x} . \bigg(\frac{1 }{1 - cosx} - \frac{sinx}{1 - cosx} \bigg) dx \\ \\ \int{e}^{x} . \bigg(\frac{1 }{2 {sin}^{2} \frac{x}{2} } - \frac{2sin \frac{x}{2} cos \frac{x}{2} }{2 {sin}^{2} \frac{x}{2}} \bigg) dx \\ \\ \int{e}^{x} . \bigg(\frac{ {cosec}^{2} \frac{x}{2} }{2 } - cot\frac{x}{2} \bigg) dx \\ \\ take \: ( - ) \: common\\ \\ - \int{e}^{x} . \bigg(\frac{ - {cosec}^{2} \frac{x}{2} }{2 } + cot\frac{x}{2} \bigg) dx \\ \\ let \: f(x) = cot\frac{x}{2} \\ \\ f'(x) = \frac{ - {cosec}^{2} \frac{x}{2} }{2 } \\ \\ thus \\ \\ \bold{\int{e}^{x} . \bigg(\frac{1 }{1 - cosx} - \frac{sinx}{1 - cosx} \bigg) dx = - {e}^{x} cot \frac{x}{2} + C}$

Hope it helps you.

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1)integration of sin x/sin x+cos x dx.

https://brainly.in/question/16108539

2)integration sin x Cos x/ sin x + cos x dx

https://brainly.in/question/1769690

Answer 2

Answer:

thus

thus∫e x .( 1−cosx1 − 1−cosxssin )dx=−e x cot 2x +C )