I have the following task:
First to prove that for $x \in [0,1)$ that $1-x\leq e^{-x} \leq \frac{1}{1+x}$
Then to deduce that $\int^1_0(1-x^2)^ndx\leq\int^1_0e^{-nx^2}dx\leq\int^1_0\frac{1}{(1+x^2)^n}dx$ for all $n \geq1$
So now, using the results above which i've completed, i have to use substitutions to deduce that for all $ n\geq 1$, that $$\int^{\pi/2}_0(\cos\theta)^{2n+1}d\theta \leq \frac{1}{\sqrt{n}}\int^{\sqrt{n}}_0e^{-y^2}dy\leq\int^{\pi/4}_0(\cos\theta)^{2n-2}d\theta$$
I have made appropriate substitutions for the first 2 integrals, but i'm struggling with the third one. I know that I have to make a transformation, for which $\theta=\pi/4 \Rightarrow u=1$. So one of the transformations appropriate for this is $\tan(\theta)$. But then when I do this, I get:
$$\int^{\pi/4}_0\left(\frac{\sin(\theta)}{\tan(\theta)}\right)^{2n-2}=$$
$u=\tan(\theta) \Rightarrow du=\frac{1}{\cos^2(\theta)}d\theta$
$$=\int^1_0\frac{\sin(\theta)^{2n-1}}{u^{2n-2}}\cos^2(\theta)d\theta=\int^1_0\frac{\sin^2(\theta)^{n-1}-\sin(\theta)^{2n}}{u^{2n-2}}d\theta$$
Well, i'm not sure what to do with this intergal. I can't see how i'm supposed to get it into the form of $\int^1_0\frac{1}{(1+x^2)^n}dx$
I also used subsition $u=\frac{\sin(\theta)}{\sqrt{2}}$ but to no avail.
Is it possible that it's a typo? It's supposed to be $\pi/2$? (the upper bound of the integral)?
If anyone could help me, that would be great!
Best Answer
There's no typo, you just need a trigonometric identity. With $x=\tan\theta$ so that $1+x^2=\sec^2\theta$ (in fact this is the motive for our choice of substitution),$$\int_0^1\frac{dx}{(1+x^2)^n}=\int_0^{\pi/4}\frac{\sec^2\theta d\theta}{(\sec^2\theta)^n}=\int_0^{\pi/4}\cos^{2n-2}\theta d\theta.$$If you wanted to deduce a suitable substitution going the other way, Bioche's rules would guide you.