Find $\int_{0}^{\pi /2}\dfrac{{1-\cos}(2\theta)}{2}\,d\theta$.

Solution We use properties of integrals to simplify before integrating. $$\begin{eqnarray*} \int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}\,d\theta &=&\dfrac{1}{2}\int_{0}^{\pi /2}\left[ 1-\cos (2\theta ) \right] \,d\theta \\ &=& \dfrac{1}{2}\left[ \int_{0}^{\pi /2}d\theta -\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \right]\\ & =& \dfrac{1}{2}\int_{0}^{\pi /2}d\theta -\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{1}{2} \Big[\theta\Big] _{0}^{\pi /2}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{\pi }{4}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \end{eqnarray*}$$

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In the integral on the right, we use the substitution $u = 2\theta$. Then $du = 2\,d\theta$ so $d\theta =\dfrac{du}{2}$. Now we change the limits of integration:

Now $$\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta =\int_{0}^{\pi }{\cos u\,\dfrac{{du}}{{2}}}= \dfrac{1}{2}\Big[\sin u\Big] _{0}^{\pi }=\dfrac{1}{2}\left( \sin \pi -\sin 0\right) =0$$

Then, $$\int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}d\theta =\dfrac{\pi}{4}-\dfrac{1}{2}\int_{0}^{\pi /2}{\cos }(2\theta ) {\,d\theta }=\dfrac{\pi }{4}$$