44. $int_{0}^{\frac{pi}{2}}cos xsin(sin x)dx$

Explanation:

Step1: Use substitution

Let $u = \sin x$. Then $du=\cos xdx$. When $x = 0$, $u=\sin(0)=0$; when $x=\frac{\pi}{2}$, $u=\sin(\frac{\pi}{2}) = 1$. The integral $\int_{0}^{\frac{\pi}{2}}\cos x\sin(\sin x)dx$ becomes $\int_{0}^{1}\sin(u)du$.

Step2: Integrate $\sin(u)$

The antiderivative of $\sin(u)$ is $-\cos(u)$. So, $\int_{0}^{1}\sin(u)du=-\cos(u)\big|_{0}^{1}$.

Step3: Evaluate the definite - integral

$-\cos(u)\big|_{0}^{1}=-\cos(1)-(-\cos(0))=1 - \cos(1)$.

Answer:

$1-\cos(1)$