# How to solve this double integral involving trig substitution (using tangent function)?

This is a question I came across and I cannot find the answer.

By using a substitution involving the tangent function, show that $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}$$

My attempt

I use trig substitution, by saying $$\tan(\theta)=\frac{y}{x}$$ which means $$x\sec^2(\theta)\,d\theta=dy$$ Also, it should be noted that because of this $$x\sec(\theta)=\sqrt{x^2+y^2}$$ $$x^4\sec^4(\theta)=(x^2+y^2)^2$$ Thus, when I substitute this information into the integral, I get $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2-(x^2\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)} \, d\theta \,dx$$ Then, this simplifies to $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx$$ $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2\sec^2(\theta)}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx=\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})}{\frac{1}{x}}\,d\theta \,dx$$ which leads to $$\int_0^1 \left[\frac \theta x \right]_0^{\arctan\left(\frac{1}{x}\right)} \,dx = \int_0^1 \frac{\arctan\left(\frac{1}{x}\right)}{x} \, dx_{(3)}$$ At this point I am stuck. How do I evaluate this integral. Am I on the right path? Wolfram Alpha gives an answer other than $\frac{\pi}{4}$ for (3), so I am not sure where I am wrong.

• Perhaps this is your culprit: $1 - \tan^2\theta \ne \sec^2\theta$. – suneater May 12 '16 at 6:17

Let us start considering $$I=\int\frac{x^2-y^2}{(x^2+y^2)^2}dy$$ Defining $$y=x\tan(\theta)\implies dy=x \sec ^2(\theta )\implies I=\int \frac{\cos (2 \theta )}{x}\,d\theta=\frac{\sin (2 \theta )}{2 x}$$ Back to $x$ $$I=\frac{y}{x^2+y^2}$$ So, $$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy=\frac 1 {1+x^2}$$ So, you are left with $$\int_0^1\frac {dx} {1+x^2}$$ Repeat the same change of variable.
• how do you know that those substitutions become $$\frac{cos(2\theta)}{x}$$ – user278039 May 12 '16 at 5:28
Starting with your work... \begin{align} \int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{(1-\tan^2(\theta))}{x\sec^2(\theta)}\,d\theta \,dx \\\\ =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{(\cos^2\theta-\sin^2\theta)}{x}\,d\theta \,dx \\\\ =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{\cos2\theta}{x}\,d\theta \,dx \\\\ =&\frac12\int_0^1 \frac{\sin2\theta}{x}\bigg|_0^{\arctan(1/x)} \,dx \\\\ = &\int_0^1 \frac{\sin\theta\cos\theta}{x}\bigg|_0^{\arctan(1/x)} \,dx \\\\ =&\int_0^1 \frac{x}{x(1+x^2)} \,dx = \arctan(1)=\pi/4 \end{align}