Question

evaluate the definite integral.\\(\int_{0}^{\frac{\pi}{4}} 2e^{\sin(x)} \cos(x) dx\\)

Explanation:

Step1: Choose substitution

Let $u = \sin(x)$, so $du = \cos(x)dx$.

Step2: Adjust integral bounds

When $x=0$, $u=\sin(0)=0$; when $x=\frac{\pi}{4}$, $u=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$.

Step3: Rewrite integral

The integral becomes $\int_{0}^{\frac{\sqrt{2}}{2}} 2e^u du$.

Step4: Compute antiderivative

Antiderivative of $2e^u$ is $2e^u$.

Step5: Evaluate at bounds

$2e^{\frac{\sqrt{2}}{2}} - 2e^0 = 2e^{\frac{\sqrt{2}}{2}} - 2$.

Answer:

$2e^{\frac{\sqrt{2}}{2}} - 2$