मराठी

Find ∫ex(1-sinx1-cosx)dx.

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प्रश्न

Find `int e^x ((1 - sinx)/(1 - cosx))dx`.

बेरीज
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उत्तर

Let, I = `int e^x ((1 - sinx)/(1 - cosx))dx`

= `int e^x (1/(1 - cosx) - sinx/(1 - cosx))dx`

= `int e^x ((-sinx)/(1 - cosx) + 1/(1 - cosx))dx`

Let, f(x) = `(-sinx)/(1 - cosx)`

f'(x) = `-[(cosx(1 - cosx) - sinx (sinx))/(1 - cosx)^2]`

= `-[(cosx - cos^2x - sin^2x)/(1 - cosx)^2]`

= `-[(cosx - (cos^2x + sin^2x))/(1 - cosx)^2]`

= `-[(cosx - 1)/(1 - cosx)^2]`

= `(1 - cosx)/(1 - cosx)^2`

= `1/(1 - cosx)`

Hence, the given integration is of form

`int e^x [f(x) + f^'(x)]dx = e^x f(x)`

where f(x) = `(-sinx)/(1 - cosx)` and f'(x) = `1/(1 - cosx)`

∴ I = `e^x xx ((-sinx)/(1 - cosx))`

= `(e^x sinx)/((cosx - 1))`.

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