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  • Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 27 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 27 Maths Textbook Solution.

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Answer:  \frac{1}{2}

Given:  \int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x

Hint: Apply the formula of  \int uvdx

Solution:   I=\int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x ……… (1)

\begin{aligned} &I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} e^{x} \cos x d x \\ & \end{aligned}

{\left[\because \int u v d x=u \int v d v-\int \frac{d}{d x} u \int v d x\right]}

\begin{aligned} &\Rightarrow I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\left(e^{x} \sin x\right)_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x \\ & \end{aligned}

\Rightarrow I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\left(e^{x} \sin x\right)_{0}^{\frac{\pi}{4}}-I[\therefore \text { from }(1)]

\begin{aligned} &\Rightarrow 2 I=-e^{\frac{\pi}{4}} \times \frac{1}{\sqrt{2}}-(-1)+e^{\frac{\pi}{4}} \times \frac{1}{\sqrt{2}}-0 \\ & \end{aligned}

\Rightarrow 2 I=1 \Rightarrow I=\frac{1}{2}

 

 

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