Question

The value of definite integral $\underset{0}{\overset{\frac{\pi }{12}}{\int }}\frac{{\tan }^{2}x-3}{3{\tan }^{2}x-1}dx$ is equal to

• A. $\frac{\pi }{12}+\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}-1}{2-\sqrt{3}}\right)$
• B. $\frac{\pi }{8}+\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}-1}{2-\sqrt{3}}\right)$
• C. $\frac{\pi }{12}-\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}-1}{2-\sqrt{3}}\right)$
• D. $\frac{\pi }{8}-\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}-1}{2-\sqrt{3}}\right)$

Answer 1

Answer: (A)

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
So $I=\underset{0}{\overset{\frac{\pi }{12}}{\int }}\frac{{\tan }^{2}x-3}{3{\tan }^{2}x-1}dx=\underset{0}{\overset{2-\sqrt{3}}{\int }}\frac{t^2-3}{\left(3t^2-1\right)\left(1+t^2\right)}dt$
$=\underset{0}{\overset{2-\sqrt{3}}{\int }}\left(\frac{1}{1+t^2}-\frac{2}{3t^2-1}\right)dt=\underset{0}{\overset{2-\sqrt{3}}{\int }}\frac{dt}{1+t^2}-\frac{2}{3}\underset{0}{\overset{2-\sqrt{3}}{\int }}\frac{dt}{t^2-{\left(\frac{1}{\sqrt{3}}\right)}^2}$
$=\frac{\pi }{12}+\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}-1}{2-\sqrt{3}}\right)$