Verified Solution • 2025–26 Session moderate

QJEEM19Maef9e: JEE Main 2019 Mathematics - Class 12 JEEM19Maef9e Solution

The value of ∫0π∣cos⁡x∣3 dx\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx is :

Multiple Choice Options

  • A 434 \over 3
  • B −- 434 \over 3
  • C 0
  • D 232 \over 3

Step-by-Step Verified Solution

∫0π∣cos⁡x∣3 dx\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx

The period of ∣cos⁡x∣\left| {\cos x} \right| = π2{\pi \over 2}

∴\therefore   I = 2 ∫0π2∣cos⁡x∣3 dx\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx

as in the range 0 to π2{\pi \over 2} ∣cos⁡x∣\left| {\cos x} \right| is positive.

So,      ∣cos⁡x∣\left| {\cos x} \right| = cosxcosx

∴\therefore   I = 2 ∫0π2cos⁡3x dx\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx}

= 2 ∫0π2(cos⁡3x+3cos⁡x4)dx\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx

I = 12[sin⁡3x3+3sin⁡x]0π2{1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}

I = 12[13(3π2)+3.sin⁡π2]{1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]

I = 12[−13+3]{1 \over 2}\left[ { - {1 \over 3} + 3} \right]

= 12(83){1 \over 2}\left( {{8 \over 3}} \right)

= 43{4 \over 3}

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