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

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

Answers (1)

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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