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

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

Answers (1)

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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