QJEEM12Ma7126: JEE Main 2012 Mathematics - Class 12 JEEM12Ma7126 Solution: If the then is equal to :

Question

Verified Solution • 2025–26 Session moderate

If the

\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,}

then

a

is
equal to :

Accepted Answer

\int {{{5\tan x} \over {\tan x - 2}}} dx

= \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx

= \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx

= \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}}

= \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx

= \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx

= \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx

= \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx

= \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx

= {I_1} + {I_2}

where

{I_1} = \int {dx}

and

{I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx}

Put

\sin x - 2\cos x = t

\Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt

\therefore

{I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C

= 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C

Hence,

{I_1} + {I_2}

= \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c}

= x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k

\Rightarrow a = 2

Step-by-Step Verified Solution