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Evaluate:

\(\rm \int _0^{\pi\over6}x\sin3x^2dx\)


1. \(\rm {\cos {\pi\over12}\over6}\)
2. \(\rm {1-\cos {\pi\over12}\over6}\)
3. \(\rm {1-\cos {\pi^2\over12}\over6}\)
4. \(\rm {\cos {\pi^2\over12}\over6}\)

1 Answer

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Correct Answer - Option 3 : \(\rm {1-\cos {\pi^2\over12}\over6}\)

Concept:

Integral property:

  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C
  • ∫ edx = ex+ C
  • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

  • A new variable is to be chosen, say “t”
  • The value of dt is to be determined.
  • Substitution is done and the integral function is then integrated.
  • Finally, the initial variable t, to be returned.

 

Calculation:

I = \(\rm \int x\sin3x^2dx\)

Let 3x2 = t; ⇒ 6x dx = dt

I = \(\rm {1\over6}\int \sin tdt\)

I = \(\rm -{1\over6} \cos t\)

I = \(\rm -{1\over6}\cos 3x^2\)

Putting up the limits
I = \(\rm \left[-{1\over6}\cos 3x^2\right]_0^{\pi\over6}\)

I = \(\rm -{1\over6}\left[\cos 3({\pi\over6})^2 - \cos 3(0)^2\right]\)

I = \(\rm -{1\over6}\left[\cos {\pi^2\over12} - 1\right]\)

I = \(\rm {1-\cos {\pi^2\over12}\over6}\)

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