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Question 227 of 255

Q.Evaluate: ∫ex[cos⁡x−sin⁡xsin⁡2x]dx\displaystyle\int e^x\left[\dfrac{\cos x - \sin x}{\sin^2 x}\right]dx.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 2mImportance★★★★★
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Recognise the integrand in the form ex[f(x)+f′(x)]e^x[f(x)+f'(x)] with f(x)=−csc⁡xf(x)=-\csc x, using ∫ex[f(x)+f′(x)] dx=exf(x)+c\int e^x[f(x)+f'(x)]\,dx = e^xf(x)+c.

Rewrite the integrand:

ex[cos⁡x−sin⁡xsin⁡2x]=ex[cos⁡xsin⁡2x−1sin⁡x]e^x\left[\frac{\cos x-\sin x}{\sin^2x}\right] = e^x\left[\frac{\cos x}{\sin^2 x} - \frac{1}{\sin x}\right]

Let f(x)=−1sin⁡x=−csc⁡xf(x) = -\dfrac{1}{\sin x} = -\csc x. Then:

f′(x)=csc⁡xcot⁡x=1sin⁡x⋅cos⁡xsin⁡x=cos⁡xsin⁡2xf'(x) = \csc x\cot x = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = \frac{\cos x}{\sin^2 x}

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