Evaluating $\int_{\pi/4}^{\pi/2}{(2\csc(x))^{17}dx}$

Question

The integral $$\int_{\pi/4}^{\pi/2}{(2\csc(x))^{17}dx}$$ is equal to:

a)$\int_0^{ln(1+\sqrt2)}{2(e^u+e^{-u})^{16}du}$

b)$\int_0^{ln(1+\sqrt2)}{2(e^u+e^{-u})^{17}du}$

c)$\int_0^{ln(1+\sqrt2)}{2(e^u-e^{-u})^{16}du}$

d)$\int_0^{ln(1+\sqrt2)}{2(e^u-e^{-u})^{17}du}$

---

My Attempt

The $(e^u - e^{-u})$ term just begs us to use $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

So $$I=\int_{\pi/4}^{\pi/2}{\frac{2^{34}i}{(e^{ix}-e^{-ix})^{17}}dx}$$

Let

$u=ix$

$du=idx$

$$I=\int_{i\pi/4}^{i\pi/2}{\frac{2^{34}}{(e^{u}-e^{-u})^{17}}du}$$

The limits of the integral tell me that I'm not on the right track. Any help to evaluate the integral(preferably using complex numbers) is appreciated.

Answer 1

$$I=\int_{\pi/4}^{\pi/2} (2 \csc x)^{17} dx$$ Let $$\csc x+\cot x=e^{u}, \csc x- \cot x= e^{-u} \implies 2 \csc x=(e^{u}+ e^{-u}), 2 \cot x=e^u- e^{-u},$$ also, $$-2\cot x \csc x dx=(e^u+e^{-u}) du \implies dx=-\frac{2du}{(e^u+e^{-u})}$$ $$x=\pi/4 \implies u=\ln (1+\sqrt{2}), ~~x=\pi/2 \implies u=0$$ Then, $$I =\int_{0}^{\ln(1+\sqrt{2})} 2(e^u+e^{-u})^{16}du.$$ Finally, $(a)$ option is correct.

Answer 2

As @Quantum noted, substituting $du=\csc xdx$ proves a) is correct. Expressing $\csc x$ in terms of this $u$ doesn't use complex eponentials, despite their role in their formula for $\sin x$.

Bioche's rules suggest instead substituting $v=\cos x$, giving $\int_0^{1/\sqrt{2}}2^{17}(1-v^2)^{-9}dv$. While this doesn't easily relate to the original problem (it doesn't help you think to try $u=\tfrac12\ln\tfrac{1+v}{1-v}$), it does make a role for complex exponentials look even less likely, because of an obvious (very tedious) integral evaluation strategy over $\Bbb R$ in partial fractions.

Why does this problem not call for complex exponentials, even though trigonometry is definable in terms of it? The simplest answer I can give is that there's actually a real-only way to relate real exponentials to complex ones, although you can rewrite it in terms of complex numbers if you really want to. Define the Gudermannian function$$\operatorname{gd}x:=2\arctan\tanh\tfrac{x}{2}$$and its inverse the Lambertian$$\operatorname{lam}x:=2\operatorname{artanh}\tan\tfrac{x}{2},$$odd functions which satisfy (among other things)$$\operatorname{gd}^\prime x=\operatorname{sech}x,\,\operatorname{lam}^\prime x=\sec x.$$If it seems weird that this works, notice$$t=\tan\frac{y}{2}=\tanh\frac{z}{2}\implies\sin y=\tanh z=\frac{2t}{1+t^2},\,\cos y=\operatorname{sech}z=\frac{1-t^2}{1+t^2}.$$Famously, integrating odd powers of the (co)secant reduces to the Lambertian (whereas with their hyperbolic counterparts the Gudermannian comes up). So the large odd power of $17$ hides the fact that you may as well be asking why integrant $\csc x$ doesn't use complex exponentials.