Let I=∫tan⁡θdθ Put tan⁡θ=x2sec2⁡θdθ=2xdxdθ=2xdx1+tan2⁡θdθ=2x⋅dx1+x4I=∫x22xdx1+x4=∫2x21+x4dx=∫x2+x2+1−11+x4dx=∫x2+1x4+1dx+∫x2−1x4+1dx Nr \& Dr by x2 in both integrals ∵∫1+1x2x2+1x2dx+∫1−1x2x2+1x2dx−eq(1) Let I1=∫1+1x2x2+1x2dx=∫1+1x2x−1x2+(2)2dx Put x−1x=u1+1x2dx=duI1=∫duu2+(2)2=12tan−1⁡u2=12tan−1⁡x−1x2+c1 Now I2=∫1−1x2x+1x2−(2)2dx Put x+1x=v1−1x2dx=dvI2=∫dvv2−(2)2=122log⁡v−2v+2+c=122log⁡x−1x−2x−1x+2+c2=122log⁡x2−2x−1x2+2x+1+c2 Substituting I1&I2 in (1) I=12tan−1⁡x2−12x+12log⁡x2−2x−1x2+2x+1+c=12tan−1⁡tan⁡θ−12tan⁡θ+122log⁡tan⁡θ−2tan⁡θ−1tan⁡θ+2tan⁡θ+1+c

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$\int \sqrt{\tan θ} d θ =$

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$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) + \frac{1}{2 \sqrt{2}} \log |\frac{\tan θ - \sqrt{2 \tan θ} + 1}{\tan θ + \sqrt{2 \tan θ} + 1}| + C$

$\sec θ - \tan θ + c$

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{\tan θ} - 1}{2 \tan θ}) + \frac{1}{2 \sqrt{2}} \log |\frac{\tan θ - \sqrt{2} \tan θ + 1}{\tan θ + \sqrt{2} \tan θ + 1}| + c$

$\frac{1}{2 \sqrt{\tan θ}} + c$

answer is A.

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Detailed Solution

$\begin{array}{l} Let I = \int \sqrt{\tan θ} d θ \\ Put \tan θ = x^{2} \\ \sec^{2} θ d θ = 2 x d x \\ d θ = \frac{2 x d x}{1 + \tan^{2} θ} \\ d θ = \frac{2 x \cdot d x}{1 + x^{4}} \\ I = \int \sqrt{x^{2}} \frac{2 x d x}{1 + x^{4}} \\ = \int \frac{2 x^{2}}{1 + x^{4}} d x \end{array}$

$\begin{array}{l} = \int \frac{x^{2} + x^{2} + 1 - 1}{1 + x^{4}} d x \\ = \int \frac{x^{2} + 1}{x^{4} + 1} d x + \int \frac{x^{2} - 1}{x^{4} + 1} d x \\ Nr \& D r by x^{2} in both integrals \\ ∵ \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x + \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x - eq (1) \\ Let I_{1} = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x \\ = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + (\sqrt{2})^{2}} d x \\ Put x - \frac{1}{x} = u \end{array}$

$\begin{array}{l} (1 + \frac{1}{x^{2}}) d x = d u \\ I_{1} = \int \frac{d u}{u^{2} + (\sqrt{2})^{2}} \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) + c_{1} \\ Now I_{2} = \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - (\sqrt{2})^{2}} d x \\ Put x + \frac{1}{x} = v \end{array}$

$\begin{array}{l} (1 - \frac{1}{x^{2}}) d x = d v \\ I_{2} = \int \frac{d v}{v^{2} - (\sqrt{2})^{2}} \\ = \frac{1}{2 \sqrt{2}} \log |\frac{v - \sqrt{2}}{v + \sqrt{2}}| + c \\ = \frac{1}{2 \sqrt{2}} \log |\frac{x - \frac{1}{x} - \sqrt{2}}{x - \frac{1}{x} + \sqrt{2}}| + c_{2} \\ = \frac{1}{2 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x - 1}{x^{2} + \sqrt{2} x + 1}| + c_{2} \\ Substituting I_{1} & I_{2} in (1) \end{array}$

$\begin{array}{l} I = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) + \frac{1}{\sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x - 1}{x^{2} + \sqrt{2} x + 1}| + c \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) + \frac{1}{2 \sqrt{2}} \log |\frac{\tan θ - \sqrt{2 \tan θ} - 1}{\tan θ + \sqrt{2 \tan θ} + 1}| + c \end{array}$