I need to calculate this integral:
$\int_0^\pi\frac {dx}{1+\sin^2x}$
I know how to solve $\int\frac {dx}{1+\sin^2x}$, I did it by substitute $t=\operatorname{cot}(x)$. But here in order to do the substitution I need to calculate the new limits by $t=\operatorname{cot}(0), t=\operatorname{cot}(\pi)$, but $\operatorname{cot}$ is not defined in $0$ and $\pi$ so I don't know how to find the new limits.
Any help will be appreciated!
On
Here's a very ridiculous and ill-advised approach; throughout, I'll sweep technical details under the rug, but rest assured Euler probably wouldn't care so maybe you shouldn't care too much either.
Expanding as a power series, we get $$\int_{0}^{\pi} \frac{1}{1+\sin^2 (x)} \, dx = \int_{0}^{\pi} \left(\sum_{k=0}^{\infty} (-1)^{k} \sin^{2k} (x) \right) \, dx = \sum_{k=0}^{\infty} (-1)^{k} \left( \int_{0}^{\pi} \sin^{2k} (x) \, dx \right) = \sum_{k=0}^{\infty} (-1)^k a_k$$ where we have set $a_{k} = \int_{0}^{\pi} \sin^{2k} (x) \,dx$.
Let us compute a generating function for this sequence $\{a_k\}$. The standard "power-reduction" formula (you can get it from integrating by parts) gives $$a_k = \frac{2k-1}{2k} a_{k-1} = \left(1 - \frac{1}{2k} \right) a_{k-1}$$
Define the function $$f(x) = \sum_{k=0}^{\infty} \frac{a_k}{k+1} x^{k+1}$$ Note that $f'(x) = \sum_{k=0}^{\infty} a_k x^k$, which is the usual generating function, and the value of the sum in question is $f'(-1)$. The recurrence gives us the relation $2(a_k - a_{k-1}) = -a_{k-1}/k$, so we have $$\frac{f(x)}{x} = \sum_{k=0}^{\infty} \frac{a_k}{k+1} x^k = \sum_{k=0}^{\infty} -2(a_{k+1} - a_k) x^k = 2\left( f'(x) - \frac{f'(x) - a_0}{x}\right)$$
Now, we have to solve the differential equation (since $a_0 = \pi$) $$f(x) = 2(\pi + xf'(x)-f'(x) ) \implies f(x) - 2\pi = 2(x-1)f'(x)$$ which we can solve easily by separation of variables. This yields (using the initial condition $f(0) = 0$) the function $$f(x) = 2\pi(1 - \sqrt{1-x})$$ In our final act of brazen recklessness, we ignore all issues of convergence and evaluate $$f'(-1) = \frac{\pi}{\sqrt{2}}$$ Miraculously, we have obtained the right answer (which you can check by evaluating the integral in a sane way, e.g. a trig sub).
On
I don't know if you like residue method. In fact \begin{eqnarray} \int_0^\pi\frac{1}{1+\sin^2x}dx&=&\int_0^\pi\frac{1}{1+\frac{1-\cos(2x}{2}}dx\\ &=&\int_0^{\pi}\frac{2}{3-\cos(2x)}dx\\ &=&\int_0^{2\pi}\frac{1}{3-\cos(x)}dx\\ &=&\int_{|z|=1}\frac{1}{3-\frac{z+\frac1z}{2}}\frac{1}{iz}dz\\ &=&\frac1i\int_{|z|=1}\frac{2}{6z-z^2-1}dz\\ &=&\frac1i\cdot2\pi i\text{Re}(\frac{1}{6z-z^2-1},3-2\sqrt2)\\ &=&2\pi\cdot\frac2{4\sqrt2}\\ &=&\frac{\pi}{\sqrt2}. \end{eqnarray}
On
One way to avoid generalized integrals is to use the symmetry of cosine and sine more. This is done in the calculation below (in the step where the primitive is calculated, one could think of doing the substitution $u=\tan x$, in the last step we use the fact that $\arctan\sqrt{2}+\arctan(1/\sqrt{2})=\pi/2$). $$ \begin{aligned} \int_0^\pi\frac{1}{1+\sin^2x}\,dx&=2\int_0^{\pi/2}\frac{1}{1+\sin^2x}\,dx\\ &=\int_0^{\pi/2}\biggl(\frac{1}{1+\sin^2x}+\frac{1}{1+\cos^2x}\biggr)\,dx\\ &=2\int_0^{\pi/4}\biggl(\frac{1}{1+\sin^2x}+\frac{1}{1+\cos^2x}\biggr)\,dx\\ &=2\biggl[\frac{1}{\sqrt{2}}\arctan(\tan(x)/\sqrt{2})+\frac{1}{\sqrt{2}}\arctan(\sqrt{2}\tan x)\biggr]_0^{\pi/4}\\ &=\frac{\pi}{\sqrt 2}. \end{aligned} $$
The integral can still be evaluated with the substitution $x\mapsto\cot x$. First, multiply the numerator and denominator by $\csc^2x$ to get$$I=\int\limits_0^{\pi}dx\,\frac {\csc^2x}{\csc^2x+1}=\int\limits_0^{\pi}dx\,\frac {\csc^2x}{\cot^2x+2}$$Now substitute $x\mapsto\cot x$. The lower limit becomes $+\infty$ while the upper limit becomes $-\infty$. Hence$$\begin{align*}I & =-\int\limits_{\infty}^{-\infty}dx\,\frac 1{x^2+2}=\int\limits_0^{\infty}dx\,\frac 1{\left(\frac x{\sqrt2}\right)^2+1}\end{align*}$$The remaining integral can be evaluated in terms of the inverse tangent function, and the final answer comes out to be