Question

$I:\int \frac{1}{3+4\cos x}dx=\frac{1}{\sqrt{7}}log\left[\frac{\sqrt{7}+\tan \left(x/2\right)}{\sqrt{7}-\tan \left(x/2\right)}\right]+c$

$II:\int \frac{dx}{\cos x-\sin x}=\frac{1}{\sqrt{2}}\log |\tan (\frac{x}{2}-\frac{3\pi }{8})|+c$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II are true

Answer 1

The correct option is C both I and II are true

$1:\int \frac{1}{3+4\cos x}dxt=\tan \frac{x}{2}=\frac{1}{2}{\sec }^2\frac{x}{2}dx=dtI:\int \frac{2dt}{[3+4\left(\frac{1-t^2}{1+t^2}\right)](1+t^2)}=\int \frac{2dt}{3(1+t^2)+4(1-t^2)}=\int \frac{2dt}{7-t^2}=\frac{2}{\sqrt{7}}\left[{\tan }^{-1}\left(\frac{t}{\sqrt{7}}\right)\right]=\frac{1}{\sqrt{7}}\log \left[\frac{\sqrt{7}+\tan \frac{x}{2}}{\sqrt{7}-\tan \frac{x}{2}}\right]+C2:\int \frac{dx}{\cos x-\sin x}=\frac{1}{\sqrt{2}}\log \left[\tan (\frac{x}{2}-\frac{3\pi }{8})\right]+Cr=\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}$

$\sin x-\cos x=\sqrt{2}\sin (x+\theta )=\sin (x-\frac{\pi }{4})=\sin (x+\theta )x=\frac{-\pi }{4}or\frac{3\pi }{4}$

$\therefore \int \frac{dx}{\cos x\sin x}=\frac{1}{\sqrt{2}}\log \left[\tan \frac{1}{2}(x-\frac{3\pi }{4})\right]+C=\frac{1}{\sqrt{2}}\log \left[\tan (\frac{x}{2}-\frac{3\pi }{8})\right]+C$