Definite Integral from 0 to Half Pi of x over Sine x

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Theorem

$\ds \int_0^{\pi/2} \frac x {\sin x} \rd x = 2 G$

where $G$ is Catalan's constant.


Proof

From Definite Integral from $0$ to $1$ of $\dfrac {\arctan x} x$, we have:

$\ds \int_0^1 \frac {\arctan x} x \rd x = G$

Let:

$x = \tan \theta$

By Derivative of Tangent Function, we have:

$\ds \frac {\d x} {\d \theta} = \sec^2 \theta$

We have, by Arctangent of Zero is Zero:

as $x \to 0$, $\theta \to 0$.

We also have, by Arctangent of One:

as $x \to 1$, $\theta \to \dfrac \pi 4$

We therefore have:

\(\ds \int_0^1 \frac {\arctan x} x \rd x\) \(=\) \(\ds \int_0^{\pi/4} \frac {\sec^2 \theta \map \arctan {\tan \theta} } {\tan \theta} \rd \theta\) substituting $x = \tan \theta$
\(\ds \) \(=\) \(\ds \int_0^{\pi/4} \frac \theta {\cos^2 \theta} \times \frac {\cos \theta} {\sin \theta} \rd \theta\) Definition of Real Secant Function, Definition of Real Tangent Function
\(\ds \) \(=\) \(\ds \int_0^{\pi/4} \frac \theta {\sin \theta \cos \theta} \rd \theta\)
\(\ds \) \(=\) \(\ds \int_0^{\pi/4} \frac \theta {\frac 1 2 \sin 2 \theta} \rd \theta\) Double Angle Formula for Sine
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^{\pi/2} \frac \phi {\sin \phi} \rd \phi\) substituting $\phi = 2 \theta$

giving:

$\ds \int_0^{\pi/2} \frac \phi {\sin \phi} \rd \phi = 2 G$

$\blacksquare$


Sources

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