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

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MarkFL

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Feb 24, 2012
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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.
 
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MarkFL

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Feb 24, 2012
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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:

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