CBSE Class 12 Mathematics 2023: question 31(a)

Hints

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Step-by-step solution

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1. Let $I=\int e^x\sin xdx$. Taking $\sin x$ as the first function: $I=e^x\sin x-\int e^x\cos xdx$.

   1 mark

2. Integrating by parts again: $I=e^x\sin x-\left[e^x\cos x+\int e^x\sin xdx\right]=e^x\sin x-e^x\cos x-I$.

   ½ mark

3. So $2I=e^x(\sin x-\cos x)$, i.e. $I=\frac{1}{2}{e}^x(\sin x-\cos x)$.

   ½ mark

4. ${\int }_0^{\pi /2}{e}^x\sin xdx=\frac{1}{2}\left[e^{\pi /2}(1-0)-e^0(0-1)\right]=\frac{1}{2}\left(e^{\pi /2}+1\right)$

   1 mark

Answer: $\frac{1}{2}\left(e^{\pi /2}+1\right)$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)