Question

The value of the integral $\int \frac{dx}{3\sin x+4\cos x}$
(where $C$ is integration constant)

• A. $\frac{1}{15}\ln \left|\frac{1-2\tan \frac{x}{2}}{4+2\tan \frac{x}{2}}\right|+C$
• B. $\frac{1}{5}\ln \left|\frac{2+\tan \frac{x}{2}}{4-2\tan \frac{x}{2}}\right|+C$
• C. $\frac{1}{5}\ln \left|\frac{1+2\tan \frac{x}{2}}{4-2\tan \frac{x}{2}}\right|+C$
• D. $\frac{1}{5}\ln \left|\frac{1-2\tan \frac{x}{2}}{4+2\tan \frac{x}{2}}\right|+C$

Answer 1

The correct option is C $\frac{1}{5}\ln \left|\frac{1+2\tan \frac{x}{2}}{4-2\tan \frac{x}{2}}\right|+C$

Given : $\int \frac{dx}{3\sin x+4\cos x}$
Now, assuming
$t=\tan \frac{x}{2}\Rightarrow x=2{\tan }^{-1}t\Rightarrow dx=\frac{2dt}{1+t^2}\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$

Now, the integral becomes
$\int \frac{dx}{3\sin x+4\cos x}=\int \frac{\frac{2dt}{1+t^2}}{3\cdot \frac{2t}{1+t^2}+4\cdot \frac{1-t^2}{1+t^2}}=\int \frac{2dt}{6t+4-4t^2}=\frac{2}{4}\int \frac{dt}{1-(t^2-\frac{6}{4}t)}=\frac{1}{2}\int \frac{dt}{1-(t^2-\frac{6}{4}t+\frac{9}{16}-\frac{9}{16})}=\frac{1}{2}\int \frac{dt}{1+\frac{9}{16}-{(t-\frac{3}{4})}^2}=\frac{1}{2}\int \frac{dt}{{\left(\frac{5}{4}\right)}^2-{(t-\frac{3}{4})}^2}=\frac{1}{2}\cdot \frac{1}{2\cdot \frac{5}{4}}\ln \left|\frac{\frac{5}{4}+t-\frac{3}{4}}{\frac{5}{4}-t+\frac{3}{4}}\right|+C=\frac{1}{5}\ln \left|\frac{\frac{1}{2}+t}{2-t}\right|+C=\frac{1}{5}\ln \left|\frac{1+2t}{4-2t}\right|+C=\frac{1}{5}\ln \left|\frac{1+2\tan \frac{x}{2}}{4-2\tan \frac{x}{2}}\right|+C$