Computing Definite Integrals — Question 2

Question 2 - Solution

Split the integral: $$\int_{0}^{\pi/2}\sin x\,dx + \int_{0}^{\pi/2}\sin^2 x\,dx.$$

Evaluate the first integral: $$\int_{0}^{\pi/2}\sin x\,dx = \bigl[-\cos x\bigr]_{0}^{\pi/2} = 1.$$

For the second integral, use the identity $$\sin^2 x=\frac{1-\cos(2x)}{2}.$$

Then $$\int_{0}^{\pi/2}\sin^2 x\,dx = \frac12\int_{0}^{\pi/2}(1-\cos 2x)\,dx.$$

Integrate: $$\frac12\left[x-\frac12\sin 2x\right]_{0}^{\pi/2} = \frac12\left(\frac{\pi}{2}-0\right) = \frac{\pi}{4}.$$

Combine results: $$1+\frac{\pi}{4}.$$

$$\boxed{1+\frac{\pi}{4}}$$