Substitution Rule for Definite Integrals — Question 5

Question 5 - Solution

This integral is well suited for substitution because the derivative of $\sin x$ appears in the numerator.

Let $$u=1+\sin x.$$ Then $$du=\cos x\,dx.$$

Change the limits of integration. When $x=0$, $\sin 0=0$, so $u=1$. When $x=\frac{\pi}{2}$, $\sin\!\left(\frac{\pi}{2}\right)=1$, so $u=2$.

Substitute into the integral: $$\int_{0}^{\pi/2} \frac{\cos x}{1+\sin x}\,dx = \int_{1}^{2} \frac{1}{u}\,du.$$

Integrate: $$\int \frac{1}{u}\,du=\ln u.$$

Apply the limits: $$\ln u\Big|_{1}^{2} = \ln 2.$$

$$\boxed{\ln 2}$$