Question
$\text{Evaluate}\int\limits^{\pi}_{0} e^{2x} . \sin\big(\frac{\pi}{4} + x\big)\text{dx}$
✓
Answer
$\text{Let I } = \int^{\pi}_{0}\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)e^{2x}\text{dx}$
$ = \sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg]^{\pi}_{0} -\int^{\pi}_{0} \cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}\text{dx}$
$\text{I} = \Bigg[\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}- \frac{1}{2}\Bigg(\cos\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg)\Bigg]^{\pi}_{0} + \frac{1}{2}\int^{\pi}_{0}-\sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{x}\text{dx}$
$\frac{5}{4}\text{I} =\bigg\{\frac{1}{2}\bigg[2\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)-\cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\bigg]e^{2x}\bigg\}^{\pi}_{0}$
$\text{I} = \frac{1}{5}\bigg\{2\bigg(-\frac{1}{\sqrt{2}}\bigg)+ \frac{1}{\sqrt{2}}\bigg\}e^{2\pi} - \bigg\{2\bigg(\frac{1}{\sqrt{2}}\bigg)-\frac{1}{\sqrt{2}}\bigg\}\Bigg] = \frac{-1}{5\sqrt{2}}(e^{2x + 1})$
$ = \sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg]^{\pi}_{0} -\int^{\pi}_{0} \cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}\text{dx}$
$\text{I} = \Bigg[\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}- \frac{1}{2}\Bigg(\cos\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg)\Bigg]^{\pi}_{0} + \frac{1}{2}\int^{\pi}_{0}-\sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{x}\text{dx}$
$\frac{5}{4}\text{I} =\bigg\{\frac{1}{2}\bigg[2\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)-\cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\bigg]e^{2x}\bigg\}^{\pi}_{0}$
$\text{I} = \frac{1}{5}\bigg\{2\bigg(-\frac{1}{\sqrt{2}}\bigg)+ \frac{1}{\sqrt{2}}\bigg\}e^{2\pi} - \bigg\{2\bigg(\frac{1}{\sqrt{2}}\bigg)-\frac{1}{\sqrt{2}}\bigg\}\Bigg] = \frac{-1}{5\sqrt{2}}(e^{2x + 1})$
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