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

AI Thread Summary
The definite integral of arccos(tan(x)) from -π/4 to π/4 can be computed using a substitution method. By letting w = tan(x), the integral transforms into I = ∫ from -1 to 1 of arccos(w)/(w² + 1) dw. Integration by parts is applied, leading to the evaluation of boundary terms and the recognition that the remaining integral is odd, resulting in a value of zero. Consequently, the final result for the integral is I = (π/2)², approximately 2.4674. This method highlights a clever approach to handling complex integrals.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$I=0\cdot\frac{\pi}{4}-\pi\left(-\frac{\pi}{4} \right)=\left(\frac{\pi}{2} \right)^2\approx2.46740110027234$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top