Verified Solution • 2025–26 Session moderate

QJEEM06Ma1a26: JEE Main 2006 Mathematics - Class 12 JEEM06Ma1a26 Solution

∫−3π2−π2[(x+π)3+cos⁡2(x+3π)]dx\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx is equal to

Multiple Choice Options

  • A π432{{{\pi ^4}} \over {32}}
  • B π432+π2{{{\pi ^4}} \over {32}} + {\pi \over 2}
  • C π2{\pi \over 2}
  • D π4−1{\pi \over 4} - 1

Step-by-Step Verified Solution

I=∫−3π2−π2[(x+π)3+cos⁡2(x+3π)] dxI = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx

Put x+π=tx + \pi = t

I=∫−π2π2(t3+cos⁡2t)dtI = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt}

=2∫−π2π2cos⁡2tdt= 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt

[\left[ {} \right. using the property of even and odd function ]\left. {} \right]

=∫0π2(1+cos⁡2t)dt=π2+0= \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0

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