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The value of $$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$ is :

JEE · Math · previous-year question

  1. A.$$4 \over 3$$correct
  2. B.$$-$$ $$4 \over 3$$
  3. C.0
  4. D.$$2 \over 3$$

Answer

A. $$4 \over 3$$

Explanation

$$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$ The period of $$\left| {\cos x} \right|$$ = $${\pi \over 2}$$ $$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$$ as in the range 0 to $${\pi \over 2}$$ $$\left| {\cos x} \right|$$ is positive. So, $$\left| {\cos x} \right|$$ = $$cosx$$ $$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $$ = 2$$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$$ I = $${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$$ I = $${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$$ I = $${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$$ = $${1 \over 2}\left( {{8 \over 3}} \right)$$ = $${4 \over 3}$$

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