evaluate the definite integral.\n int_{0}^{\frac{pi}{4}} sin(4t)dt

Explanation:

Step1: Use substitution

Let $u = 4t$, then $du=4dt$ and $dt=\frac{1}{4}du$. When $t = 0$, $u = 0$; when $t=\frac{\pi}{4}$, $u=\pi$. The integral becomes $\frac{1}{4}\int_{0}^{\pi}\sin(u)du$.

Step2: Integrate $\sin(u)$

The antiderivative of $\sin(u)$ is $-\cos(u)$. So, $\frac{1}{4}\int_{0}^{\pi}\sin(u)du=\frac{1}{4}[-\cos(u)]_{0}^{\pi}$.

Step3: Evaluate the definite - integral

$\frac{1}{4}[-\cos(u)]_{0}^{\pi}=\frac{1}{4}[-\cos(\pi)-(-\cos(0))]=\frac{1}{4}[-(-1)-(-1)]=\frac{1}{4}(1 + 1)=\frac{1}{2}$.

Answer:

$\frac{1}{2}$