MHB Calculate Definite Integral of arcos(tanx) from -pi/4 to pi/4

[MarkFL]

Here is the question:

Calculate this definite integral?


Definite integral of arcos(tanx) dx from -pi/4 to pi/4
I know this isn't an easy antiderivative but my professor said there was a easy trick to compute this nonetheless.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Jeremy,

We are given to evaluate:

$$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^{-1}\left(\tan(x) \right)\,dx$$

Consider the following substitution:

$$w=\tan(x)\,\therefore\,dw=\sec^2(x)\,dx$$

But, if we square the substitution and apply a Pythagorean identity, we find:

$$w^2=\tan^2(x)=\sec^2(x)-1\implies \sec^2(x)=w^2+1$$

And so we may state:

$$dx=\frac{1}{w^2+1}\,dw$$

And so our definite integral becomes:

$$I=\int_{-1}^{1} \frac{\cos^{-1}(w)}{w^2+1}\,dw$$

Applying integration by parts, let:

$$u=\cos^{-1}(w)\,\therefore\,du=-\frac{1}{\sqrt{1-w^2}}\,dw$$

$$dv=\frac{1}{w^2+1}\,dw\,\therefore\,v=\tan^{-1}(w)$$

Hence, we may state:

$$I=\left[\cos^{-1}(w)\tan^{-1}(w) \right]_{-1}^{1}+\int_{-1}^{1} \frac{\tan^{-1}(w)}{\sqrt{1-w^2}}\,dw$$

Now, observing that the remaining integrand is odd and the limits symmetric, we know it's value is zero, and so we are left with:

$$I=0\cdot\frac{\pi}{4}-\pi\left(-\frac{\pi}{4} \right)=\left(\frac{\pi}{2} \right)^2\approx2.46740110027234$$