Question 249 of 416MediumMCQ
The value of ∫0π∣cosx∣3dx is:
Solutions · 1
Official solution
\[\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx\]
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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