Question: (\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x} dx\)…

\(\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x} dx\)

Solution

To solve the integral

\[ \int_0^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x} \, dx \]

we can use the substitution method.

Let \( u = 1 + \sin x \).

Then, \( du = \cos x \, dx \).

When \( x = 0 \), \( u = 1 + \sin 0 = 1 \).

When \( x = \frac{\pi}{2} \), \( u = 1 + \sin \frac{\pi}{2} = 2 \).

The integral becomes

\[ \int_1^{2} \frac{1}{u} \, du \]

Integrating \( \frac{1}{u} \), we get

\[ \ln |u| \]

Thus, the integral evaluates to

\[ \left[ \ln |u| \right]_1^2 \]

Substituting the limits:

\[ \ln |2| - \ln |1| = \ln 2 - 0 = \ln 2 \]

Therefore, the value of the integral is \( \ln 2 \).