Browse Questions

# Evaluate: $\large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\; $$dx \begin{array}{1 1}(A) \sin^{-1} (\tan x) + k \\ (B) \cos^{-1} (\tan x) + k \\ (C) \sin^{-1} (\cot x) + k \\ (D) \cos^{-1} (\sec x) + k \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • 1 + \tan^2 x = \sec^2 x • \large \int \frac{1}{\sqrt(1-x^2)}\;$$dx =\sin^{-1} x + k$
• $\large \frac{d(\tan x)}{dx}$$=\sec^{2} x Given: \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx$
Let $\tan x = t \rightarrow \sec^2x\;dx = dt$
We know that: $1 + \tan^2 x = \sec^2 x$
Substituting, we get: $\large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \large \int \frac {\sec^2x\;dx}{\sqrt(1 - \tan^2x)} \Rightarrow \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \large \int \frac{dt}{\sqrt (1-t^2)}$
We know that: $\large \int \frac{1}{\sqrt(1-x^2)}\;$$dx =\sin^{-1} x + k \Rightarrow \large \int \frac {1 + \tan^2 x} {\sqrt (1 - \tan^2 x)}\;$$dx = \sin^{-1}(t) + k$