Definite Trigonometry exercise

Exercise

Compute, $\int_{-\frac{\pi }{2}}^{\pi }{\sqrt{2\sin x\sin 2x+4{{\sin }^{2}}x}\,dx}$

Solution: we know that $\sin 2x=2\sin x\cos x\Leftrightarrow 2\sin x\sin 2x=2\sin x\left( 2\sin x\cos x \right)=4{{\sin }^{2}}x\cos x$

Thus $\sqrt{2\sin x\sin 2x+4{{\sin }^{2}}x}=\sqrt{4{{\sin }^{2}}x\cos x+4{{\sin }^{2}}x}=\sqrt{4{{\sin }^{2}}x\left( 1+\cos x \right)}$

So $\int_{-\frac{\pi }{2}}^{\pi }{\sqrt{2\sin x\sin 2x+4{{\sin }^{2}}x}\,dx}=\int_{\frac{-\pi }{2}}^{\pi }{\sqrt{4{{\sin }^{2}}x\left( 1+\cos x \right)}\,dx}=\int_{\frac{-\pi }{2}}^{\pi }{\left| 2\sin x \right|\sqrt{1+\cos x}\,dx}$

Let $u=\cos x\Leftrightarrow \,du=-\sin x\,dx$ so $\forall x\in \left[ \frac{-\pi }{2},0 \right]\,,\sin x<0\,\,$ and $\sin x>0\,\,\forall x\in \left[ 0,\pi  \right]$

So $\int_{\frac{-\pi }{2}}^{\pi }{\left| 2\sin x \right|\sqrt{1+\cos x}}\,dx=\int_{\frac{-\pi }{2}}^{0}{-2\sin x\sqrt{1+\cos x}\,dx}+\int_{0}^{\pi }{2\sin x\sqrt{1+\cos x}\,dx}$

Hence $\int_{\frac{-\pi }{2}}^{0}{-2\sin x\sqrt{1+\cos x}\,dx}=2\int_{u\left( -\frac{\pi }{2} \right)}^{u\left( 0 \right)}{\sqrt{1+u}\,du}=2\int_{0}^{1}{\sqrt{1+u}\,d}\left( 1+u \right)$

$=\left. \frac{4}{3}{{\left( 1+u \right)}^{3/2}} \right|_{0}^{1}=\left. \frac{4}{3}\sqrt{{{\left( 1+u \right)}^{3}}} \right|_{0}^{1}=\frac{4}{3}\left( \sqrt{{{2}^{3}}}-1 \right)=\frac{4}{3}\left( 2\sqrt{2}-1 \right)$

And $\int_{0}^{\pi }{2\sin x\sqrt{1+\cos x}\,dx}=-2\int_{u\left( 0 \right)}^{u\left( \pi  \right)}{\sqrt{1+u}\,du}=-2\int_{1}^{-1}{\sqrt{1+u}\,d\left( 1+u \right)}$

$=\left. \frac{-4}{3}{{\left( 1+u \right)}^{3/2}} \right|_{1}^{-1}=-\frac{4}{3}\left. \sqrt{{{\left( 1+u \right)}^{3}}} \right|_{1}^{-1}=\frac{-4}{3}\left( 0-2\sqrt{2} \right)=\frac{8\sqrt{2}}{2}$

Thus $\int_{\frac{-\pi }{2}}^{\pi }{\sqrt{2\sin \sin 2x+4{{\sin }^{2}}x}\,dx}=\frac{8\sqrt{2}}{3}-\frac{4}{3}+\frac{8\sqrt{2}}{3}=\frac{16\sqrt{2}-4}{3}$