Substitution Rule for Definite Integrals — Question 2

Question 2 - Solution

This integral suggests a substitution involving $\cos x$ because its derivative appears in the integrand.

Let $$u=1+\cos x.$$ Then $$du=-\sin x\,dx \quad\Rightarrow\quad -\!du=\sin x\,dx.$$

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

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

Integrate: $$\int \sqrt{u}\,du = \frac{2}{3}u^{3/2}.$$

Apply the limits: $$\frac{2}{3}u^{3/2}\Big|_{1}^{2} = \frac{2}{3}\left(2^{3/2}-1\right).$$

$$\boxed{\frac{2}{3}\left(2^{3/2}-1\right)}$$