Question

$\int e^{\frac{x}{\sqrt{2}}}\cos \left(\frac{x}{\sqrt{2}}\right)dx=.....+c$

• A. $e^{\frac{x}{\sqrt{2}}}\sin (\frac{x}{\sqrt{2}}-\frac{\pi }{4})$
• B. $e^{\frac{x}{\sqrt{2}}}\sin (\frac{x}{\sqrt{2}}+\frac{\pi }{4})$
• C. $e^{\frac{x}{\sqrt{2}}}\cos (\frac{x}{\sqrt{2}}+\frac{\pi }{4})$
• D. $\frac{e^{\frac{x}{\sqrt{2}}}}{\sqrt{2}}\sin (\frac{x}{\sqrt{2}}-\frac{\pi }{4})$

Answer 1

Answer: (B)

$e^{\frac{x}{\sqrt{2}}}\sin (\frac{x}{\sqrt{2}}+\frac{\pi }{4})$

Consider, $I=\int e^{\frac{x}{\sqrt{2}}}\cos \left(\frac{x}{\sqrt{2}}\right).dx$

Let

$\frac{x}{\sqrt{2}}=t$ $\Rightarrow$ $x=\sqrt{2}t$ $\Rightarrow$ $dx=\sqrt{2}.dt$

$\Rightarrow$ $I=\int e^t-\cos t\sqrt{2}.dt$

$=\sqrt{2}\int e^t.\cos t.dt$

$=\sqrt{2}[\cos t\int e^t.dt-\int \frac{d}{dt}\cos t.\int e^t.dt]$

$=\sqrt{2}[\cos t.e^t+\int \sin t.e^t.dt]$

$=\sqrt{2}[\cos t.e^t+\sin t\int e^t-\int \frac{d}{dt}\sin t.\int e^t.dt].$

$=\sqrt{2}(\cos t.e^t+\sin t.e^t-\int \cos t.e^t.dt)$

$=(1-\sqrt{2})\int e^{\frac{x}{2}\cos \left(\frac{x}{\sqrt{2}dt}\right)}=\sqrt{2}(\cos t.e^t+\sin t.dt)$

$=\int e^{\frac{x}{\sqrt{2}}}.\cos \left(\frac{x}{\sqrt{2}}\right).dt=\frac{\sqrt{2}}{1-\sqrt{2}}.\frac{x}{\sqrt{2}}(\cos t+\sin t)+c$

$=\frac{x}{e^{\sqrt{2}}}.\sin (\frac{x}{\sqrt{2}}+\frac{\pi }{4})$