∫tanxa+btan2xdx =∫sin xa cos2x+b sin2xdx =∫sin xb+cos2x(a-b) Put cos x=t⇒-sin x dx=dt =∫-dtb-(b-a)t2 =-1b-a∫dtbb-a-t2 =1b-a cos-1 tbb-a+C =1b-acos-1 b-ab.cos x+C

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$\int \frac{\tan x}{\sqrt{a + b \tan^{2} x}} d x =$

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$\frac{1}{\sqrt{b - a}} \cos^{- 1} [\sqrt{\frac{b - a}{b}} \cos x] + c$

$- \frac{1}{\sqrt{b - a}} \cos^{- 1} [\sqrt{\frac{b - a}{b}} \cos x] + c$

$\frac{2}{\sqrt{b - a}} \cos^{- 1} [\sqrt{\frac{b - a}{b}} \cos x] + c$

$- \frac{2}{\sqrt{b - a}} \cos^{- 1} [\sqrt{\frac{b - a}{b}} \cos x] + c$

answer is A.

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Detailed Solution

$\int \frac{tanx}{\sqrt{a + {btan}^{2} x}} dx = \int \frac{\sin x}{\sqrt{a \cos^{2} x + b \sin^{2} x}} dx = \int \frac{\sin x}{\sqrt{b + \cos^{2} x (a - b)}} Put \cos x = t \Rightarrow - \sin x dx = dt = \int \frac{- dt}{\sqrt{b - (b - a) t^{2}}} = \frac{- 1}{\sqrt{b - a}} \int \frac{dt}{\sqrt{\frac{b}{b - a} - t^{2}}} = \frac{1}{\sqrt{b - a}} \cos^{- 1} (\frac{t}{\sqrt{\frac{b}{b - a}}}) + C = \frac{1}{\sqrt{b - a}} \cos^{- 1} (\sqrt{\frac{b - a}{b}} . \cos x) + C$