Evaluate ∫13+sin⁡2xdx

I=∫13+sin⁡2xdx=∫13sin2⁡x+cos2⁡x+2sin⁡xcos⁡xdx=∫sec2⁡x3tan2⁡x+2tan⁡x+3dx[Dividing Nr and Dr by cos2 x] Putting tan x = t and sec2 x dx = dt, we get I=∫dt3t2+2t+3=13∫dtt2+23t+1=13∫dtt+132+2232I=131223tan−1⁡t+13223+C=122tan−1⁡3t+122+C=122tan−1⁡3tan x+122+C

Evaluate ∫13+sin⁡2xdx

I=∫13+sin⁡2xdx=∫13sin2⁡x+cos2⁡x+2sin⁡xcos⁡xdx=∫sec2⁡x3tan2⁡x+2tan⁡x+3dx[Dividing Nr and Dr by cos2 x] Putting tan x = t and sec2 x dx = dt, we get I=∫dt3t2+2t+3=13∫dtt2+23t+1=13∫dtt+132+2232I=131223tan−1⁡t+13223+C=122tan−1⁡3t+122+C=122tan−1⁡3tan x+122+C

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$Evaluate \int \frac{1}{3 + \sin 2 x} d x$

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

$\frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{\tan x + 1}{2 \sqrt{2}}) + C$

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int \frac{1}{3 + \sin 2 x} d x \\ = \int \frac{1}{3 (\sin^{2} x + \cos^{2} x) + 2 \sin x \cos x} d x \\ = \int \frac{\sec^{2} x}{3 \tan^{2} x + 2 \tan x + 3} d x \\ [D i v i d i n g N r a n d D r b y \cos^{2} x] P u t t i n g \tan x = t a n d s e c^{2} x d x = d t, w e g e t \\ \begin{array}{l} I = \int \frac{d t}{3 t^{2} + 2 t + 3} = \frac{1}{3} \int \frac{d t}{t^{2} + \frac{2}{3} t + 1} \\ = \frac{1}{3} \int \frac{d t}{{(t + \frac{1}{3})}^{2} + {(\frac{2 \sqrt{2}}{3})}^{2}} \\ I = \frac{1}{3} \frac{1}{(\frac{2 \sqrt{2}}{3})} \tan^{- 1} (\frac{\frac{t + \frac{1}{3}}{2 \sqrt{2}}}{3}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{3 t + 1}{2 \sqrt{2}}) + C \\ = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{3 \tan x + 1}{2 \sqrt{2}}) + C \end{array} \end{array}$