# Definite Trigonometric Integration

Calculate the integral : $$\ \int_{0}^{\pi}\frac{x}{a-\sin(x)}dx , \quad a>1$$

I have tried trig substitutions, and replacing $\ x$ with $\ \pi-y$, along with other commonly used "tricks", but none seem to work.

• Attempting to integrate with Mathematica, this integral seems rather complicated. Are you sure you copied it down correctly. e.g. polylogarithms feature in the indefinite integral, amongst other various inverse trigonometric functions... I don't think simple substitutions are going to work here. Integrating w.r.t $a$ is easier, but that's not what the current question is asking... – Pixel Jan 8 '18 at 15:38
• You must have overlooked something, because "replacing $\ x$ with $\ \pi-y$" does work. – Professor Vector Jan 8 '18 at 15:39
• The $x\leftrightarrow\pi-x$ trick seems to be on the right track. It establishes $$\int_0^{\pi}\frac{xdx}{a-\sin x}=\frac{\pi}{2}\int_0^{\pi}\frac{dx}{a-\sin x}.$$ From mathematical experience, if you can factor out a $\pi$ and the remaining looks simpler, it is simpler. Yeah, now mathematica has the answer for the right-hand side. – Zhuoran He Jan 8 '18 at 15:40

As mentioned in the comments, the substitution $x \to \pi - x$ yields :

$$\int_0^{\pi}\frac{xdx}{a-\sin x}=\frac{\pi}{2}\int_0^{\pi}\frac{dx}{a-\sin x}$$

I'll go over the calculation of the integral on the RHS.

We have :

$$I ={\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x =-{\displaystyle\int}\dfrac{1}{\sin\left(x\right)-a}\,\mathrm{d}x$$

Solving the integral : $$I= {\displaystyle\int}\dfrac{1}{\sin\left(x\right)-a}\,\mathrm{d}x$$

Let's prepare our integral for the Weierstrass Substitution :

$${\displaystyle\int}\dfrac{1}{\frac{2\tan\left(\frac{x}{2}\right)}{\tan^2\left(\frac{x}{2}\right)+1}-a}\,\mathrm{d}x$$

Substitute :

$$u=\tan\left(\dfrac{x}{2}\right)\to \mathrm{d}x=\dfrac{2}{\sec^2\left(\frac{x}{2}\right)}\,\mathrm{d}u =\dfrac{2}{u^2+1}\,\mathrm{d}u$$

The integral then becomes :

$$I=-\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{au^2-2u+a}\,\mathrm{d}u = -{\displaystyle\int}\dfrac{1}{\left(\sqrt{a}u-\frac{1}{\sqrt{a}}\right)^2+a-\frac{1}{a}}\,\mathrm{d}u$$

Substitute :

$$v=\dfrac{au-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}} \to \mathrm{d}u=\dfrac{\sqrt{a-\frac{1}{a}}}{\sqrt{a}}\,\mathrm{d}v$$

The integral then becomes :

$$I=-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}}}{\displaystyle\int}\dfrac{1}{v^2+1}\,\mathrm{d}v = -\dfrac{\arctan\left(v\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}}$$

Substituting back, we get :

$$I=-\dfrac{\arctan\left(\frac{au-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}} = \dots =-\dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}}$$

The integral is now solved, as by :

$$-{\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x =\dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}}$$

Which by simplifying and rewriting, leads to :

$$\boxed{{\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x= \dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}}+C, \quad a>1}$$

Can you now, using this result and the result from the substitution, calculate the given indefinite integral :

$$\ \int_{0}^{\pi}\frac{x}{a-\sin(x)}dx , \quad a>1$$