Question

1. evaluate the following integral:

\int \cos x \ln(\sin x) dx

Explanation:

Step1: Substitution

Let \( u = \sin x \), then \( du = \cos x \, dx \). The integral becomes \( \int \ln(u) \, du \).

Step2: Integration by parts

For \( \int \ln(u) \, du \), let \( v = \ln(u) \), \( dw = du \). Then \( dv = \frac{1}{u} du \), \( w = u \). By integration by parts formula \( \int v \, dw = vw - \int w \, dv \), we get \( u\ln(u) - \int u \cdot \frac{1}{u} du = u\ln(u) - \int 1 \, du \).

Step3: Evaluate remaining integral

\( \int 1 \, du = u + C \), so \( u\ln(u) - u + C \).

Step4: Substitute back

Replace \( u \) with \( \sin x \), we have \( \sin x \ln(\sin x) - \sin x + C \).

Answer:

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