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Find The Exact Value of A Definite Integral

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anemone

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Feb 14, 2012
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Find the exact value of the following definite integral:

\(\displaystyle \int_0^2 (3x^2-3x+1)\cos (x^3-3x^2+4x-2)\,dx\)
 

MarkFL

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Feb 24, 2012
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Interesting problem! I noticed neither my calculator (TI-89) nor W|A could come up with the exact result. Here is my approach:

Let:

\(\displaystyle I=\int_0^2\left(3x^2-3x+1 \right)\cos\left(x^3-3x^2+4x-2 \right)\,dx\)

Observe that:

\(\displaystyle x^3-3x^2+4x-2=(x-1)^3+(x-1)\)

\(\displaystyle 3x^2-3x+1=3x(x-1)+1\)

So, let:

\(\displaystyle u=x-1\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+3u+1 \right)\cos\left(u^3+u \right)\,du\)

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+1 \right)\cos\left(u^3+u \right)\,du+3\int_{-1}^{1}u\cos\left(u^3+u \right)\,du\)

By the odd-function rule, this reduces to:

\(\displaystyle I=\int_{-1}^{1}\left(3u^2+1 \right)\cos\left(u^3+u \right)\,du\)

Let:

\(\displaystyle v=u^3+u\,\therefore\,dv=\left(3u^2+1 \right)\,du\)

and we have (using the even-function rule):

\(\displaystyle I=2\int_0^2\cos(v)\,dv=2\sin(2)\approx1.8185948536513634\)
 
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anemone

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Feb 14, 2012
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What a great solution, MarkFL!(Sun)
 

Prove It

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Jan 26, 2012
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Find the exact value of the following definite integral:

\(\displaystyle \int_0^2 (3x^2-3x+1)\cos (x^3-3x^2+4x-2)\,dx\)
I have a slightly different approach to Mark's...

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x^2 - 3x + 1 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \,dx } \\ = \int_0^2{ \left( 3x^2 - 6x + 4 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx} + \int_0^2{ \left( 3x - 3 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \,dx } \end{align*}\)

The first integral can be solved using \(\displaystyle \displaystyle \begin{align*} u = x^3 - 3x^2 + 4x - 2 \implies du = \left( 3x^2 - 6x + 4 \right) \, dx \end{align*}\) giving

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x^2 - 6x + 4 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx} &= \int_{-2}^2{ \cos{(u)}\,du } \\ &= \left[ \sin{(u)} \right]_{-2}^2 \\ &= \sin{(2)} - \sin{(-2)} \\ &= \sin{(2)} + \sin{(2)} \\ &= 2\sin{(2)} \end{align*}\)

As for the second

\(\displaystyle \displaystyle \begin{align*} \int_0^2{ \left( 3x - 3 \right) \cos{ \left( x^3 - 3x^2 + 4x - 2 \right) } \, dx } &= 3 \int_0^2{ \left( x - 1 \right) \cos{ \left[ \left( x - 1 \right) \left( x^2 - 2x + 2 \right) \right] } \, dx } \\ &= 3\int_0^2{ \left( x - 1 \right) \cos{ \left\{ \left( x - 1 \right) \left[ \left( x - 1 \right) ^2 + 1 \right] \right\} } \, dx } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} v = x - 1 \implies dv = dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} 3\int_{-1}^1{ v \cos{ \left[ v \left( v^2 + 1 \right) \right] } \, dv } &= 0 \end{align*}\)

because it's an odd function integrated over an even region.


Therefore, the original integral equals \(\displaystyle \displaystyle \begin{align*} 2\sin{(2)} \end{align*}\)