How do you evaluate the definite integral #int3xsin(x^2)# from #[0,sqrtpi]#?

Question

How do you evaluate the definite integral #int3xsin(x^2)dx# from #[0,sqrtpi]#?

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Answer 1 · 2016-09-15T18:28:59

#3.#

We know that, #intsin y dy=-cos y+C#.

Let #y=x^2," so that, "dy=2xdx#

#:. intsin x^2*2xdx=2intxsinx^2dx=-cosx^2+C#.

#:. intxsinx^2dx=-1/2cosx^2+C#

#rArr int3xsinx^2dx=3intxsinx^2dx=-3/2cosx^2+C#

Therefore,

#int_0^(sqrtpi) 3xsinx^2dx=-3/2[cosx^2]_0^(sqrtpi)#

#=-3/2[cospi-cos0]#

#=-3/2[-1-1]#

#=-3/2*(-2)#

#=3.#