Problems regarding the ratio of two definite integrals These are two interesting problems:
(1) Find the ratio of $\int_0^1 (1-t^4)^{-0.5} \,dt$ and $\int_0^1 (1+t^4)^{-0.5} \,dt$
(2) Find the ratio of $\int_0^x e^{xt-t^2} \,dt$ and $\int_0^x e^{\frac{-t^2}4} \,dt$
Both are not supposed to be solved via solving individual integrals.
I also know that the answer to (2) is $e^{\frac{x^2}4}$, but how it is arrived I am confused.
Furthermore, what's the general way of solving problems like these, the fraction of two definite integrals with same upper and lower limits?
I thank in advance any help I receive!
 A: *

*Use the change of variable formula, respectively $t=\sin u$ and $t=\tan u$:
\begin{align}
 &\int_0^1{\dfrac{1}{\sqrt{1-t^4}}\text{d}t}\\
 =&\int_0^{\pi /2}{\dfrac{\cos u}{\sqrt{1-\sin ^4u}}\text{d}u}\\
 =&\int_0^{\pi /2}{\dfrac{1}{\sqrt{1+\sin ^2u}}\text{d}u},
\end{align}
\begin{align}
 &\int_0^1{\dfrac{1}{\sqrt{1+t^4}}\text{d}t}\\
 =&\int_0^{\pi /4}{\dfrac{\sec ^2u}{\sqrt{1+\tan ^4u}}\text{d}u}\\
 =&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^4u+\cos ^4u}}\text{d}u}\\
 =&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^4u+\cos ^4u+2\sin ^2u\cos ^2u-2\sin ^2u\cos ^2u}}\text{d}u}\\
 =&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\left( \sin ^2u+\cos ^2u \right) ^2-\frac{1}{2}\sin ^22u}}\text{d}u}\\
 =&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^22u+\cos ^22u-\frac{1}{2}\sin ^22u}}\text{d}u}\\
 =&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\frac{1}{2}+\frac{1}{2}\cos ^22u}}\text{d}u}\\
 =&\int_0^{\pi /2}{\dfrac{1}{2\sqrt{\frac{1}{2}+\frac{1}{2}\cos ^2u}}\text{d}u}\\
 =&\int_0^{\pi /2}{\dfrac{1}{\sqrt{2\left( 1+\sin ^2u \right)}}\text{d}u}.
\end{align}
The ratio is $\sqrt{2}$.

*Use the change of variable formula $t-\frac x2=\frac u2$, thus $\text{d}t=\frac 12\text{d}u$:
\begin{align}
 &\int_0^x{\text{e}^{xt-t^2}\text{d}t}\\
 =&\int_0^x{\text{e}^{\frac{x^2}{4}}\text{e}^{-\frac{x^2}{4}+xt-t^2}}\text{d}t\\
 =&\text{e}^{\frac{x^2}{4}}\int_0^x{\text{e}^{-\left( t-\frac{x}{2} \right) ^2}}\text{d}t\\
 =&\text{e}^{\frac{x^2}{4}}\int_{-x}^x{\text{e}^{-\frac{u^2}{4}}\dfrac{1}{2}\text{d}u}\\
 =&\text{e}^{\frac{x^2}{4}}\int_0^x{\text{e}^{-\frac{u^2}{4}}\text{d}u}.
\end{align}
