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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following definite integrals:$\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
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Answer

Let $\text{I}=\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$ Integrating by parts, we get$\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Now, integrating the second term by parts, we get$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\bigg\{\Big[\frac{1}{2}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\Big]^{\pi}_0\\+\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\bigg\}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{4}\Big[\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^{\pi}_0-\frac{1}{4}\text{I}$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[\text{e}^{2\pi}\sin\Big(\pi+\frac{\pi}{4}\Big)-\sin\Big(\frac{\pi}{4}\Big)\Big]\\-\frac{1}{4}\Big[\text{e}^{2\pi}\cos\Big(\pi+\frac{\pi}{4}\Big)-\cos\Big(\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]-\frac{1}{4}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]$
$\Rightarrow\frac{5}{4}\text{I}=-\frac{1}{2\sqrt{2}}\text{e}^{2\pi}-\frac{1}{2\sqrt{2}}+\frac{1}{4\sqrt{2}}\text{e}^{2\pi}+\frac{1}{4\sqrt{2}}$
$\Rightarrow\text{I}=-\frac{1}{5\sqrt{2}}\big(\text{e}^{2\pi}+1\big)$

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