# Find The Exact Value of A Definite Integral

#### anemone

##### MHB POTW Director
Staff member
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

Staff member
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):

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

#### anemone

##### MHB POTW Director
Staff member
What a great solution, MarkFL!

#### Prove It

##### Well-known member
MHB Math Helper
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*}