# How to evaluate this Trig integral?

I need to find the definite integral of $\int(1+x^2)^{-4}~dx$ from $0$ to $\infty$ .

I rewrite this as $\dfrac{1}{(1+x^2)^4}$ .

The $\dfrac{1}{1+x^2}$ part, from $0$ to $\infty$ , seems easy enough: $\arctan(\infty)-\arctan(0)$ which gives $\dfrac{\pi}{2}$ .

How do I deal with the "$^4$" part of the equation.

I know the answer is $\dfrac{5\pi}{32}$ . What steps am I missing?

Thank you.

-
Along with all the answers, you can also use simple fractions. – GPerez Feb 9 '14 at 23:46

Substitute $x=\tan{t}$. Then, $$\int_0^\infty \frac{dx}{(x^2+1)^4} dx=\int_0^\frac{\pi}{2} \cos^6{t} dt=\int_0^{\frac{\pi}{2}} (\frac{1+\cos{2t}}{2})^3 dt.$$ By repeated use of formula $\cos^2{t}=\frac{1+\cos{2t}}{2}$, it is easy to get result $\frac{5\pi}{32}$.

-
I would mention that, while $\tan t$ has discontinuities at $x=\pi(2k+1)/2, k=0,1,2,\dots$, this set has null measure and doesn't affect the integral. – GPerez Feb 9 '14 at 23:45

Let $\forall n\in \mathbb{N}, I_n(x):=\int_0^x \frac{dt}{(t^2+1)^n}$ by integration by parts we get : $$I_n(x)=\frac{x}{(1+x^2)^n}+2n(I_n(x)-I_{n+1}(x))$$ Thus, $$I_{n+1}(x)=\frac{2n-1}{2n}I_n(x)+\frac{1}{2n}\frac{x}{(1+x^2)^n}$$ Note that $I_1=arctan (x)$.

Therefore, $$I_{4}(x)=\frac{1}6\frac{x}{(1+x^2)^3}+\frac{5}{24}\frac{x}{(1+x^2)^2}+\frac{5}{16}\frac{x}{(1+x^2)}+\frac{5}{16}arctan x$$ Take the limit as $x\rightarrow +\infty$ we get the result $\frac{5\pi}{32}$.

-

Consider the sub. $x=\tan\theta$ you will get a function of $\cos \theta$ with even power, try to reduce power by double angle rule

-


-

You can't deal with the integrand in that manner -- the 4th power and the $1/(1+x^2)$ must be dealt with 'together'. Denote your integral by $I$ and use a substitution of $x=\tan \theta$ and the identity $\sec ^2 \theta = 1+ \tan ^2 \theta$ to reduce the integrand to $\cos ^6\theta$.

From there, you have multiple ways in. One of which is to let $J_n = \int_0^{\pi/2} \cos^{2n}\theta d\theta$ and derive an reduction formula via parts by writing the integrand as $\cos^{2n-1}\theta \cdot \sin \theta$.

Alternatively, use Alan's approach.

EDIT: If you want a more fancy way that abuses symmetry, http://www.thestudentroom.co.uk/showthread.php?t=1608009&p=30894842#post30894842

-