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

Answers (1)

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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