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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 1 maths textbook solution

Answers (1)

Answer:
The correct answer is e^{x} \cos (x)+c
Hint:

\frac{d}{d x} \cos x=-\sin x

Given:

\int e^{x}(\cos x-\sin x) d x

Solution:

On integration by part......\int u \cdot v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]

        =\int e^{x} \cos x \; d x-\int e^{x} \sin x\; d x

We know that  \frac{d}{d x} \cos x=-\sin x 

        \begin{aligned} &=\cos x \int e^{x}-\int \frac{d}{d x} \cos x \int e^{x}-\int e^{x} \sin x \; d x \\ &=e^{x} \cos x+\int e^{x} \sin x \; d x-\int e^{x} \sin x\; d x \\ &=e^{x} \cos x+c \end{aligned}

So the answer is  e^{x} \cos x+c

 

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