# Question 7d5b1

Nov 1, 2017

$a = 0 , 1$

#### Explanation:

Notice that $x + 2 {a}^{2} - a$ is just $x$ with the added constant $2 {a}^{2} - a$. This makes the substitution $u = x + 2 {a}^{2} - a$ not too difficult, since $\mathrm{du} = \mathrm{dx}$.

We can substitute this directly into the integral, but remember that the bounds will change as we move from $x$ to $u$. The bound $x = a$ will become $u = \left(a\right) + 2 {a}^{2} - a = 2 {a}^{2}$ and the bound $x = - a$ becomes $u = \left(- a\right) + 2 {a}^{2} - a = 2 {a}^{2} - 2 a$.

We then see that:

${\int}_{- a}^{a} \cos \left(x + 2 {a}^{2} - a\right) \mathrm{dx} = {\int}_{2 {a}^{2} - 2 a}^{2 {a}^{2}} \cos \left(u\right) \mathrm{du}$

The integral of cosine is just sine:

=[sin(u)]_(2a^2-2a)^(2a^2)=color(blue)(sin(2a^2)-sin(2a^2-2a)#

Which is not equivalent to what you stated. You may have attempted to do ${\left[\sin \left(u\right)\right]}_{2 {a}^{2} - 2 a}^{2 {a}^{2}} = \sin \left(2 {a}^{2} - \left(2 {a}^{2} - 2 a\right)\right)$, but this is not how to properly evaluate the function.

You could also get a "simpler" answer using the sine difference to product formula: $\sin \left(\alpha\right) - \sin \left(\beta\right) = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$.

Here that gives the reduction:

$= 2 \cos \left(\frac{2 {a}^{2} + 2 {a}^{2} - 2 a}{2}\right) \sin \left(\frac{2 {a}^{2} - \left(2 {a}^{2} - 2 a\right)}{2}\right) = 2 \cos \left(2 {a}^{2} - a\right) \sin \left(- a\right)$

Upon re-examining this, perhaps you want a solution for $a$.

In that case, we want that $2 \cos \left(2 {a}^{2} - a\right) \sin \left(- a\right) = - \sin \left(2 a\right)$.

Note that $- \sin \left(2 a\right) = - 2 \cos \left(a\right) \sin \left(a\right)$. Also note that $2 \cos \left(2 {a}^{2} - a\right) \sin \left(- a\right) = - 2 \cos \left(2 {a}^{2} - a\right) \sin \left(a\right)$. So, the two are equal when $\cos \left(2 {a}^{2} - a\right) = \cos \left(a\right)$, which occurs when $2 {a}^{2} - a = a$.

Solving this yields $2 {a}^{2} - 2 a = 2 a \left(a - 1\right) = 0$, so $a = 0$ or $a = 1$.