$\smallint \frac{{\sin \,\,2x}}{{1\,\, + \,\,{{\cos }^2}\,\,x}}\,\,dx$ is equal to

$ - \frac{1}{2}\,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}x} \right) + c$

$2\,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}x} \right) + c$

$\frac{1}{2}\,\,\log \,\,\left( {1\,\, + \,\,\cos \,\,\,2x} \right) + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{{\sin \,\,2x}}{{1\,\, + \,\,{{\cos }^2}\,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: D

c\,\, - \,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}\,\,x} \right)

Option D,

c\,\, - \,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}\,\,x} \right)

, is correct because it is the result of integrating the given trigonometric function using a simple substitution.