Question

1. evaluate the following integral: \\(\int \cos x \ln(\sin x) dx\\)

Explanation:

Step1: Substitute $u = \sin x$

$du = \cos x dx$, integral becomes $\int \ln u du$

Step2: Integration by parts

Let $v = \ln u$, $dw = du$; then $dv = \frac{1}{u}du$, $w = u$. So $\int \ln u du = u\ln u - \int u \cdot \frac{1}{u}du = u\ln u - u + C$

Step3: Substitute back $u = \sin x$

$\sin x \ln(\sin x) - \sin x + C$

Answer:

$\sin x \ln(\sin x) - \sin x + C$