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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 11

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Answer: The required integral is \frac{-1}{5} e^{-2 x} \cos x-\frac{2}{5} e^{-2 x} \sin x+c

Hint: Let that one antiderivative of e^{-2x} has shape Ae^{-2x}\cos x+Be^{-2x}\sin x

Given: \int e^{-2 x} \sin x d x

Solution:

Let that one antiderivative of e^{-2x} has shape Ae^{-2x}\cos x+Be^{-2x}\sin x

Differentiate,we get

A\left(-e^{-2 x} \sin x-2 e^{-2 x} \cos x\right)+B\left(-e^{-2 x} \cos x-2 e^{-2 x} \sin x\right)

This is equal to

(-A-2 B) e^{-2 x} \sin x+(-2 A+B) e^{-2 x} \cos x \quad \rightarrow(1)

In order to make identically equations for A and B.

We get  A=\frac{-1}{5}e^{-2x} \: and\: B=\frac{-2}{5}

Thus our integral is \frac{-1}{5} e^{-2 x} \cos x-\frac{2}{5} e^{-2 x} \sin x+c

 

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