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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 19

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Answer:
The correct answer is e^{2 x} \cos x+c
 

Given:

\int e^{2 x}(-\sin x+2 \cos x) d x

Solution:

        I=\int e^{2 x}(-\sin x+2 \cos x) d x

            =-\int e^{2 x} \sin x \; d x+2 \int e^{2 x} \cos x\; d x

Consider   2 \int e^{2 x} \cos x \; d x

Integrating by parts, we get

\begin{aligned} &\text { Let } u=\cos x \\ &d u=-\sin x \; d x \\ &v=e^{2 x} \end{aligned}

                \int v d x=\frac{e^{2 x}}{2}

                2 \int e^{2 x} \cos x d x

            \begin{aligned} &=2\left[\int \frac{e^{2 x}}{2} \cos x-\int \frac{e^{2 x}}{2}(-\sin x) d x\right]+c \\ &=e^{2 x} \cos x+\int e^{2 x} \sin x \; d x+c \end{aligned}

\text { Now, } I=-\int e^{2 x} \sin x d x+2 \int e^{2 x} \cos x d x

              \begin{aligned} &=-\int e^{2 x} \sin x \; d x+e^{2 x} \cos x+\int e^{2 x} \sin x \; d x+c \\ &=e^{2 x} \cos x+c \end{aligned}

So, the correct answer is e^{2 x} \cos x+c

 

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