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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 29 maths

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Answer: 1

Hint: you must know the rule of integration

Given: \int_{0}^{\frac{\pi}{2}} e^{x}(\sin x-\cos x) d x

Solution:  \int_{0}^{\frac{\pi}{2}} e^{x}(\sin x-\cos x) d x

        =-\int_{0}^{\frac{\pi}{2}} e^{x}(\cos x-\sin x) d x

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

        \begin{aligned} &\text { we get, } I=-\left[e^{x} \cos x\right]_{0}^{\frac{\pi }{2}} \\\\ &=-e^{\frac{\pi}{2}} \cos \frac{\pi}{2}+e^{0} \cos (0) \\\\ &=0+(1)(1) \\\\ &=1 \end{aligned}


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