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

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

Mary Goodson · Accepted Answer

Let f(x)=4arctan⁡(ex)−π. Then since arctan⁡u+arctan⁡1u=π2 (∗):  −f(−x)=−4arctan⁡e−x+π=−4(π2−arctan⁡ex)+π=4arctan⁡ex−π=f(x)  (∗)  can be proved geometrically: arctan⁡u is the angle with opposite side u and adjacent side 1, and arctan⁡1u 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 · Accepted Answer

Write the integral as  I=∫−π04arctan⁡(ex)−πdx+∫0π4arctan⁡(ex)−πdx  Upon making the substitution u=−x, we find that  I=−∫π04arctan⁡(e−u)−πdu+∫0π4arctan⁡(ex)−πdx  =∫0π4arctan⁡(e−x)−πdx+∫0π4arctan⁡(ex)−πdx  =∫0π4(arctan⁡(ex)+arctan⁡(e−x))−2πdx  Note that  arctan⁡(u+arctan⁡(1u))={π2 if u&amp;gt;0−π2 if u&amp;lt;0  In this case, x∈[0,π], so u=ex&amp;gt;0 and arctan⁡(u)+arctan⁡(1u)=π2, meaning that the integrand is the zero function. Hence, I=0.

karton · Accepted Answer

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
$\frac{d}{d x} [4 \arctan (e^{x}) - π] = 4 \cdot \frac{1}{1 + (e^{x})^{2}} \cdot e^{x} = \frac{4}{e^{x} + e^{- x}} = 2 \sec h x$
In particular, this derivative is even. Since evaluating the integrand at x=0 gives $4 \arctan e^{0} - π = 0$ the integrand is odd; since the integral is taken over an interval symmetric around 0, by symmetry
$\int_{- a}^{a} [4 \arctan (e^{x}) - π] d x = 0$
for any a: In particular, the occurrence of $π$ in the limits of the integral is something of a red herring.

Find the value of \int_{-\pi}^{\pi} (4 \arctan(e^x)-\pi)dx I've tried to show

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