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

JEE · Math · previous-year question

  1. A.$${{{\pi ^4}} \over {32}}$$
  2. B.$${{{\pi ^4}} \over {32}} + {\pi \over 2}$$
  3. C.$${\pi \over 2}$$correct
  4. D.$${\pi \over 4} - 1$$

Answer

C. $${\pi \over 2}$$

Explanation

$$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$ Put $$x + \pi = t$$ $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $$ $$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$$ $$\left[ {} \right.$$ using the property of even and odd function $$\left. {} \right]$$ $$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$

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