QJEEM20Ma40cb: JEE Main 2020 Mathematics - Class 12 JEEM20Ma40cb Solution: If = A , where C is a constant of integration, then can be :

Question

Verified Solution • 2025–26 Session moderate

If

\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta

= A

{\log _e}\left| {B\left( \theta \right)} \right| + C

,

where C is a constant of integration, then

{{{B\left( \theta \right)} \over A}}

can be :

Accepted Answer

\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta

=

\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}}

=

\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}}

Let sin

\theta

= t

\Rightarrow

cos

\theta

d

\theta

= dt

=

\int {{{dt} \over {3 + 7t + 2{t^2}}}}

=

\int {{{dt} \over {\left( {2t + 1} \right)\left( {t + 3} \right)}}}

=

{1 \over 5}\int {\left( {{2 \over {2t + 1}} - {1 \over {t + 3}}} \right)dt}

=

{1 \over 5}\ln \left| {{{2t + 1} \over {t + 3}}} \right|

+ C

=

{1 \over 5}\ln \left| {{{2\sin \theta + 1} \over {\sin \theta + 3}}} \right|

+ C

\therefore

A =

{{1 \over 5}}

and B(

\theta

) =

{{{2\sin \theta + 1} \over {\sin \theta + 3}}}

\therefore

{{{B\left( \theta \right)} \over A}}

=

{{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}

Step-by-Step Verified Solution