The antiderivative of 1sin2⁡x+tan2⁡x is

The integrand can be written as sec2⁡xtan2⁡x2+tan2⁡xputting tan⁡x=t, the integral reduces to ∫dtt22+t2=12∫1t2−12+t2dt =12−1t−12tan−1⁡t2+C=−12cot⁡x+12tan−1⁡12tan⁡x+C

The antiderivative of 1sin2⁡x+tan2⁡x is

The integrand can be written as sec2⁡xtan2⁡x2+tan2⁡xputting tan⁡x=t, the integral reduces to ∫dtt22+t2=12∫1t2−12+t2dt =12−1t−12tan−1⁡t2+C=−12cot⁡x+12tan−1⁡12tan⁡x+C

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The antiderivative of $\frac{1}{\sin^{2} x + \tan^{2} x}$ is

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$(- 1 / 2) [\cot x + (1 / \sqrt{2}) \tan^{- 1} ((1 / \sqrt{2}) \tan x)] + C$

none of these

$- [\cot x + (1 / \sqrt{2}) \tan^{- 1} ((1 / \sqrt{2}) \tan x)]$

$(- 1 / 2) [\tan x + (1 / \sqrt{2}) \tan^{- 1} (1 / \sqrt{2}) \tan x] + C$

answer is B.

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Detailed Solution

The integrand can be written as $\frac{\sec^{2} x}{\tan^{2} x (2 + \tan^{2} x)}$

putting $\tan x = t$ , the integral reduces to

$\begin{array}{l} \int \frac{d t}{t^{2} (2 + t^{2})} = \frac{1}{2} \int (\frac{1}{t^{2}} - \frac{1}{2 + t^{2}}) d t \\ = \frac{1}{2} [- \frac{1}{t} - \frac{1}{\sqrt{2}} \tan^{- 1} \frac{t}{\sqrt{2}}] + C \\ = (- \frac{1}{2}) [\cot x + (\frac{1}{\sqrt{2}}) \tan^{- 1} ((\frac{1}{\sqrt{2}}) \tan x)] + C \end{array}$