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QJEEM18Ma84b7: JEE Main 2018 Mathematics - Class 12 JEEM18Ma84b7 Solution

The value of ∫−π/2π/2sin⁡2x1+2xdx\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx is

Multiple Choice Options

  • A π4{\pi \over 4}
  • B π8{\pi \over 8}
  • C π2{\pi \over 2}
  • D 4π{4\pi }

Step-by-Step Verified Solution

As we know,

∫ab+(x)dx=∫abf(a+b−x)dx\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx

Let I=∫−π2π2  sin⁡2x1+2x dxI = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx

⇒    I=∫−π2π2  sin⁡2(−π2+π2−x)1+2−π2+π2−x\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}}

⇒    I=∫−π2π2sin⁡2x1+2−xdx\Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx

⇒    I=∫−π2π22xsin⁡2x1+2x  dx\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx

∴   \therefore\,\,\, 2I=∫−π2π2(sin⁡2x1+2x+2xsin⁡2x1+2x)  dx2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx

⇒    2I=∫−π2π2sin⁡2x(1+2x1+2x)  dx\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx

⇒    2I=∫−π2π2sin⁡2x  dx\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx

⇒    2I=2∫0π2sin⁡2x  dx   \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\, [ as sin2 x is an even function ]

⇒    I=∫0π2sin⁡2x  dx    \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,

[ as     \,\,\,\, ∫0af(a)  dx=∫0af(a−x) dx]\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} }

So,    \,\,\, I=∫0π2sin⁡2(b2−x)  dx    I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,

   I=∫0π2cos⁡2x  dx    \,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,

∴    \therefore\,\,\,\,     2I=∫0π2(sin⁡2x+cos⁡2x)  dx    \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,

⇒    2I=∫0π2   dx    \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,

⇒    2I=[x]0π2\Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}

⇒    2I=π2\Rightarrow \,\,\,\,2I = {\pi \over 2}

⇒    I=π4\Rightarrow \,\,\,\,I = {\pi \over 4}

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