The integral $\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{ an ^2 x} d x$ is equal to :

Question

The integral $\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{	an ^2 x} d x$ is equal to :

Aicharya · Accepted Answer

Let $\mathrm{I}=\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{	an ^2 x} d x$ $ =\int_{7 \pi / 4}^{7 \pi / 3} 	an x d x=-\left.\log \cos xight|_{7 \pi / 4} ^{7 \pi / 3} $ $=-\left[\log \cos \frac{7 \pi}{3}-\log \cos \frac{7 \pi}{4}ight]$ $=\log \cos \frac{7 \pi}{4}-\log \cos \frac{7 \pi}{3}$ $=\log \left[\frac{\cos \frac{7 \pi}{4}}{\cos \frac{7 \pi}{3}}ight]=\log \left[\frac{\cos \left(2 \pi-\frac{\pi}{4}ight)}{\cos \left(2 \pi+\frac{\pi}{3}ight)}ight]$ $=\log \left(\frac{\cos \frac{\pi}{4}}{\cos \frac{\pi}{3}}ight)=\log \left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}ight)$ $=\log \left(\frac{2}{\sqrt{2}}ight)=\log \sqrt{2}$

The integral $\int_{7 \pi / 4}^{7 \pi / 3} \sqrt{\tan ^2 x} d x$ is equal to :

Solution