If ∫12+sin2xdx=k.tan−1(l tanx)+C then (k,l)=

∫12+sin2xdx=∫sec2x2sec2x+tan2xdx=∫sec2x dx3tan2x+2=∫dt3t2+2=13∫dtt2+23=1312/3tan−1t2/3+c=16tan−13tanx2+ck=16,l=32

If ∫12+sin2xdx=k.tan−1(l tanx)+C then (k,l)=

∫12+sin2xdx=∫sec2x2sec2x+tan2xdx=∫sec2x dx3tan2x+2=∫dt3t2+2=13∫dtt2+23=1312/3tan−1t2/3+c=16tan−13tanx2+ck=16,l=32

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If $\int \frac{1}{2 + \sin^{2} x} d x = k . \tan^{- 1} (l \tan x) + C$ then $(k, l) =$

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$(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{3}})$

$(\frac{1}{\sqrt{6}}, \frac{2}{3})$

$(\frac{1}{\sqrt{6}}, \sqrt{\frac{2}{3}})$

$(\frac{1}{\sqrt{6}}, \sqrt{\frac{3}{2}})$

answer is D.

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Detailed Solution

$\int \frac{1}{2 + \sin^{2} x} d x$

$= \int \frac{\sec^{2} x}{2 \sec^{2} x + \tan^{2} x} d x$

$= \int \frac{\sec^{2} x d x}{3 \tan^{2} x + 2}$

$= \int \frac{d t}{3 t^{2} + 2}$

$= \frac{1}{3} \int \frac{d t}{t^{2} + \frac{2}{3}}$

$= \frac{1}{3} \frac{1}{\sqrt{2 / 3}} \tan^{- 1} (\frac{t}{\sqrt{2 / 3}}) + c$

$= \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{\sqrt{3} \tan x}{\sqrt{2}}) + c$

$k = \frac{1}{\sqrt{6}}, l = \sqrt{\frac{3}{2}}$

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If ∫12+sin2xdx=k.tan−1(l tanx)+C then (k,l)=

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If $\int \frac{1}{2 + \sin^{2} x} d x = k . \tan^{- 1} (l \tan x) + C$ then $(k, l) =$