Evaluation of certain definite integrals How do we integrate the following two integrals? 
$$ \int_{-e}^\pi \cos(-3x^2) \, dx$$ and $$ \int_{-e}^\pi \frac {e^x}{\ln(a-x)} \, dx,$$ where $a>\pi$. 
 A: $\int_{-e}^\pi\cos(-3x^2)~dx$
$=\int_{-e}^\pi\cos3x^2~dx$
$=\int_{-e}^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^n(3x^2)^{2n}}{(2n)!}dx$
$=\int_{-e}^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^nx^{4n}}{(2n)!}dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^nx^{4n+1}}{(2n)!(4n+1)}\right]_{-e}^\pi$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^n(\pi^{4n+1}-(-e)^{4n+1})}{(2n)!(4n+1)}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^n(\pi^{4n+1}+e^{4n+1})}{(2n)!(4n+1)}$
$\int_{-e}^\pi\dfrac{e^x}{\ln(a-x)}dx$
$=\int_{a+e}^{a-\pi}\dfrac{e^{a-x}}{\ln x}d(a-x)$
$=\int_{a-\pi}^{a+e}\dfrac{e^ae^{-x}}{\ln x}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^{-e^x}}{x}d(e^x)$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^xe^{-e^x}}{x}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^ae^x}{x}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^{nx}}{n!}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^ae^{(n+1)x}}{n!x}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^a}{n!x}dx+\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$
$=\int_{\ln(a-\pi)}^{\ln(a+e)}\dfrac{e^{a-1}}{x}dx+\int_{\ln(a-\pi)}^{\ln(a+e)}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^{k-1}}{n!k!}dx$
$=\left[e^{a-1}\ln x\right]_{\ln(a-\pi)}^{\ln(a+e)}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^kx^k}{n!k!k}\right]_{\ln(a-\pi)}^{\ln(a+e)}$
$=e^{a-1}(\ln\ln(a+e)-\ln\ln(a-\pi))+\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^ne^a(n+1)^k((\ln(a+e))^k-(\ln(a-\pi))^k)}{n!k!k}$
