# Evaluate: $\rm \int _0^{\pi\over6}x\sin3x^2dx$

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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. $\rm {\cos {\pi^2\over12}\over6}$

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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 = $\rm {1-\cos {\pi^2\over12}\over6}$