Question

$\int_{}^{}\frac{1}{\sqrt{\tan^{2}\theta + 2}}$ is equal to

• A. ln$|\sqrt{\tan^{2}\theta - 1} + \sqrt{\tan^{2}\theta + 2}| + c$
• B. ln$|\sqrt{\tan^{2}\theta - 1} - \sqrt{\tan^{2}\theta + 2}| + c$
• C. ln $|\tan^{2}\theta - 1 + \sqrt{\tan^{2}\theta - 1}| + c$
• D. None of these

Answer 1

Answer: (A)

ln$|\sqrt{\tan^{2}\theta - 1} + \sqrt{\tan^{2}\theta + 2}| + c$

Answer 2

Let tan² q – 1 = t²

I = = ln | t + $\sqrt{t^{2} + 3}$| + c