QJEEM20Ma1163: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1163 Solution: If = where c is a constant of integration, then g(0) is equal…

Question

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If

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

=

g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c

where c is a constant of integration, then g(0) is equal to :

Accepted Answer

I =

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

=

\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx

=

\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx

+

\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

=

\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx

+ \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

=

\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx

+

\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

Let

{{e^x} + {e^{ - x}} + x}

= t

\Rightarrow

{\left( {{e^x} - {e^{ - x}} + 1} \right)}

dx = dt

and let

{{e^x} + {e^{ - x}}}

= u

\Rightarrow

{\left( {{e^x} - {e^{ - x}}} \right)dx}

= du

\therefore

I =

\int {{e^t}} dt + \int {{e^u}} du

=

{{e^t}}

+

{{e^u}}

+ C

=

{e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}

+

{{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}

+ C

=

{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)

+ C

\therefore

g(x) = e^x + 1

\Rightarrow

g(0) = 2

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