Question #d2780

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Answer 1 · 2017-11-06T12:07:03

$\frac{1}{2 \sqrt{3}} {tan}^{- 1} (\frac{1}{\sqrt{3}} \cdot tan 2 x) + C .$ $1/(2sqrt3)tan^-1(1/sqrt3*tan2x)+C.$

Part I :

Let, $I = \int \frac{1}{2 {cos}^{2} 2 x + 1} d x,$ $I=int1/(2cos^2 2x+1)dx,$

$= \int \frac{1}{{cos}^{2} 2 x (2 + \frac{1}{{cos}^{2} (2 x)})} d x,$ $=int1/{cos^2 2x(2+1/cos^2(2x))}dx,$

$= \int \frac{\frac{1}{{cos}^{2} (2 x)}}{2 + {sec}^{2} 2 x} d x,$ $=int(1/cos^2 (2x))/(2+sec^2 2x)dx,$

$= \int \frac{{sec}^{2} 2 x}{2 + (1 + {tan}^{2} 2 x)} d x,$ $=int(sec^2 2x)/{2+(1+tan^2 2x)}dx,$

$:. I=int(sec^2 2x)/(3+tan^2 2x)dx.$

We subst. $tan 2x=y," so that, "(sec^2 2x)(2)dx=dy.$

$:. I=int((1/2)dy)/(3+y^2),$

$=1/2*1/sqrt3*tan^-1(y/sqrt3),$

$rArr I=1/(2sqrt3)tan^-1(1/sqrt3*tan2x)+C.$

Part II :

It is not clear that what is to be integrated.