QJEEM21Ma3a07: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3a07 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx

is :

Accepted Answer

Let

I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx

.... (1)

Replace x with

-

x,

\therefore

I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}}

.... (2)

Adding (1) and (2), we get,

2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx

= \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx}

= \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx

[

{{{\cos }^2}x}

is a even function as

f(x) = f( - x)

for

{\cos ^2}x

]

= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx

= 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx

\Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx

= {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}

= {1 \over 2}\left[ {{\pi \over 2} - 0} \right]

= {\pi \over 4}

Step-by-Step Verified Solution