STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

$\therefore \int \frac{d x}{5+4 \cos x}=\int \frac{d x}{5+4\left(\frac{1-\tan ^{2}\left(\frac{\pi}{2}\right)}{1+\tan ^{2}\left(\frac{\pi}{2}\right)}\right)}$

$=\int \frac{\left(1+\tan ^{2}\left(\frac{z}{2}\right)\right) d x}{5\left(1+\tan ^{2}\left(\frac{x}{2}\right)\right)+4\left(1-\tan ^{2}\left(\frac{\pi}{2}\right)\right)}$

$=\int \frac{\sec ^{2}\left(\frac{z}{2}\right) d x}{\left(9+\tan ^{2}\left(\frac{\pi}{2}\right)\right)}$

Let $\tan \left(\frac{x}{2}\right)=y$ on differentiating this we get, $\sec ^{2}\left(\frac{x}{2}\right) \times \frac{d x}{2}=d y$

Thus, $\int \frac{\sec ^{2}\left(\frac{x}{2}\right) d x}{\left(9+\tan ^{2}\left(\frac{\pi}{2}\right)\right)}=\int \frac{2 d y}{9+y^{2}}=2 \int \frac{1}{y^{2}+3^{2}} d x=2 \times \frac{1}{3} \tan ^{-1} y+c$

Now putting in the value of y in above equation, we get,

$\int \frac{d x}{5+4 \cos x}=\frac{2}{3} \tan ^{-1}\left(\tan \frac{x}{2}\right)+c$

Then $m=\frac{2}{3}$