Find the value of ∫−ππ(4arctan⁡(ex)−π)dx Ive

Jason Yuhas

Jason Yuhas

Answered

2021-12-31

Find the value of ππ(4arctan(ex)π)dx
Ive

Answer & Explanation

Mary Goodson

Mary Goodson

Expert

2022-01-01Added 37 answers

Let f(x)=4arctan(ex)π. Then since arctanu+arctan1u=π2 (∗):
f(x)=4arctanex+π=4(π2arctanex)+π=4arctanexπ=f(x)
(∗) can be proved geometrically: arctanu is the angle with opposite side u and adjacent side 1, and arctan1u is the other acute angle in the right triangle. Their sum must thus be π2. Alternatively, differentiate the left-hand side and show it is constant, then choose a convenient value for u since any u works, say u=1.
autormtak0w

autormtak0w

Expert

2022-01-02Added 31 answers

Write the integral as
I=π04arctan(ex)πdx+0π4arctan(ex)πdx
Upon making the substitution u=−x, we find that
I=π04arctan(eu)πdu+0π4arctan(ex)πdx
=0π4arctan(ex)πdx+0π4arctan(ex)πdx
=0π4(arctan(ex)+arctan(ex))2πdx
Note that
arctan(u+arctan(1u))={π2 if u>0π2 if u<0
In this case, x[0,π], so u=ex>0 and arctan(u)+arctan(1u)=π2, meaning that the integrand is the zero function. Hence, I=0.
karton

karton

Expert

2022-01-08Added 439 answers

The answer expands on my comment, which remarked that the integrand of the given integral is twice the Gudermannian function, gd, which appears most famously in the equation governing the Mercator projection in cartography.
NSK
Differentiating the integrand gives
ddx[4arctan(ex)π]=411+(ex)2ex=4ex+ex=2sechx
In particular, this derivative is even. Since evaluating the integrand at x=0 gives 4arctane0π=0 the integrand is odd; since the integral is taken over an interval symmetric around 0, by symmetry
aa[4arctan(ex)π]dx=0
for any a: In particular, the occurrence of π in the limits of the integral is something of a red herring.

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