Is it appropriate to make a substitution $t=\tan\left(\frac{\theta}{2}\right)$ here?

Question

Suppose we have the following Riemann integral

$$\int \dfrac{1}{(\sin(x)+\cos(x))^2}dx$$

I was curious if we could make a substitution $t = \tan\biggr (\dfrac{\theta}{2}\biggr )$ here.

Answer 1

Since $$\dfrac{1}{(\sin(x)+\cos(x))^2}={1\over 2\cos^2\left(x-{\pi\over 4}\right)}$$you would better substitute$$u=\tan\left(x-{\pi\over 4}\right)$$

Answer 2

Hint:

Bioche's rules suggest to set $t=\tan x$ so $\;\mathrm dx =\dfrac{\mathrm dt}{1+t^2}$.

Indeed, expand the denominator first: $$\int\dfrac{1}{(\sin x+\cos x)^2}\,\mathrm dx=\int\dfrac{1}{1+\sin 2x}\,\mathrm dx= \int\dfrac{1}{1+\cfrac{2t}{1+t^2}}\,\frac{\mathrm dt}{1+t^2}.$$ Can you end the computation?

Answer 3

Hint:

For $1+\sin2x\ne0,$

$$\dfrac1{1+\sin2x}=\dfrac{1-\sin2x}{\cos^22x}=\sec^22x-\sec2x\tan2x$$

But in general for $$\int\dfrac{a+b\sin2x+c\cos2x}{d+e\sin2x+f\cos2x}\ dx$$

it's easier to use Weierstrass Substitution $\tan x=u$

Answer 4

You should specify the range of $x$ because depending on this $\tan \frac{x}{2}$ may be undefined and then we definitely cannot make this substitution.
As others have already pointed out, even if $\tan \frac{x}{2}$ is defined on your interval, it is not necessarily the best substitution to compute this integral. So the fact that we can make this substitution doesn't imply that we should.
A little side note: What you wish to compute isn't a Riemann integral, but an antiderivative.

Answer 5

Let's expand on @Bernard's recommendation to follow Bioche's third rule.

A rule of thumb for integrating periodic functions that occasionally doesn't work, but certainly suffices here, is to use a substitution $t=\tan kx$ for which $t$ has the same period as the integrand. The reasoning is that integrating a periodic function, minus its average over a period, over one period gives $0$, and $[f(\tan kx)]_X^{X+\pi/k}=0$ (this may require some kind of regularization for certain values of $X$.) Thus, for example, $\int_0^\pi\sin 2xdx\stackrel{t:=\tan x}{=}\int_0^0\frac{2tdt}{(1+t^2)^2}=0$. Motivating the rest of Bioche's rules is a worthwhile exercise.

While $\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$ has period $2\pi$, $(\sin x+\cos x)^k$ has period $\pi$ for even $k$, including the case at hand, $k=-2$. But $\pi$ is also the period of $\tan x$, so it's a better choice for $t$ than $\tan\frac{x}{2}$.