# Jeremy's question at Yahoo! Answers: A definite integral with the inverse cosine of the tangent

#### MarkFL

##### Administrator
Staff member
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

##### Administrator
Staff member
Hello Jeremy,

We are given to evaluate:

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

Consider the following substitution:

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

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

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

And so we may state:

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

And so our definite integral becomes:

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

Applying integration by parts, let:

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

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

Hence, we may state:

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

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:

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