# Show equivalence of different approaches to $\int\frac{\sin(x)dx}{1+\cos^2(x)}$

Given:
For the indefinite integral $$J = \displaystyle\int\dfrac{\sin(x)}{1+\cos^2(x)}\,dx,\,$$ one anti-derivative will be obtained via the substitution $$\,u=\cos(x)\,$$ and another anti-derivative will be obtained via the substitution $$\,u=\tan(x/2)\,$$. While the two anti-derivatives will appear to be incompatible, their equivalence will be explicitly shown via the generic definite integral $$\displaystyle\int_a^b\dfrac{\sin(x)}{1+\cos^2(x)}\,dx.$$

To Do:
As I will clarify at the end of this query, I am looking for an alternative (i.e. more implicit) way of demonstrating the equivalence of the two anti-derivatives, within a "constant of integration," other than evaluating the (generic) definite integral.

$$\underline{u=\cos(x)}$$
$$du = -\sin(x)dx \;\Rightarrow\; J \;=\; \displaystyle{\int} \dfrac{-du}{1+u^2}\, \;=\; -\arctan(u) \;=\; -\arctan[\cos(x)]$$
$$=\; \arctan[-\cos(x)].$$

$$\underline{u=\tan(x/2)}$$
$$x = 2\arctan(u), \;dx = \dfrac{2du}{1+u^2}, \;\sin(x) = \dfrac{2u}{1+u^2}, \;\cos(x)=\dfrac{1-u^2}{1+u^2}.$$

Under this substitution, $$J$$ will simplify to $$\,\displaystyle\int\dfrac{2u\,du}{1+u^4} \;=\; \arctan\left[\dfrac{-1}{u^2}\right]$$
$$= \;\arctan\left[\dfrac{-1}{\tan^2(x/2)}\right] \;=\; \arctan\left[\dfrac{-\cos^2(x/2)}{\sin^2(x/2)}\right] \;=\; -\arctan\left[\dfrac{1+\cos(x)}{1-\cos(x)}\right]\,.$$

$$\underline{\text{Definite integral for} \;u=\cos(x)}$$

$$\displaystyle J = \int_a^b\dfrac{\sin(x)dx}{1+\cos^2(x)}\,dx.\;$$ Let $$\;r_1 = \arctan[cos(a)]\;$$ and let $$\;s_1 = \arctan[cos(b)].$$

Then $$\;J \;=\; (r_1 - s_1).$$

$$\underline{\text{Definite integral for} \;u=\tan(x/2)}$$

$$\displaystyle J = \int_{\tan(a/2)}^{\tan(b/2)}\dfrac{2u\,du} {1+u^4}.\;$$ Let $$\;r_2 = \arctan\left[\dfrac{1}{\tan^2(a/2)}\right]\;$$ and let

$$\;s_2 = \arctan\left[\dfrac{1}{\tan^2(b/2)}\right].$$

Then $$\;J \;=\; (r_2 - s_2).$$

Equivalence of the two substitutions:
I will use the identity $$\;\tan (\alpha - \beta) \;=\; \dfrac{\tan(\alpha) - \tan(\beta)} {1 + \tan(\alpha)\tan(\beta)}\;$$ to demonstrate that
$$\;\tan(r_1 - s_1) \;=\; \tan(r_2 - s_2).$$
Although not rigorous, this strongly suggests that angle $$(r_1 - s_1) =\;$$ angle $$\;(r_2 - s_2).$$

$$\tan(r_1 - s_1) = \dfrac{\cos(a) - \cos(b)}{1 + \cos(a)\cos(b)}.$$

$$\tan(r_2 - s_2) = \dfrac {\dfrac{1+\cos(a)}{1-\cos(a)} - \dfrac{1+\cos(b)}{1-\cos(b)}} {1 + \left(\dfrac{1+\cos(a)}{1-\cos(a)}\right) \left(\dfrac{1+\cos(b)}{1-\cos(b)}\right)} \;=\; \dfrac{2[\cos(a) - \cos(b)]}{2[1+\cos(a)\cos(b)]}$$
$$=\; \tan(r_1 - s_1).$$

The real question:

Focusing on the indefinite integrals only (ignoring the definite integrals), the two anti-derivatives are $$\;-\arctan[cos(x)]\;$$ and $$\;-\arctan\left[\dfrac{1+\cos(x)}{1-\cos(x)}\right].$$ How does one demonstrate that these two anti-derivatives are (somehow) essentially the same?

• $$\frac{\pi}{4}+\arctan y=\arctan 1+\arctan y=\arctan \frac{1+y}{1-y}$$ Since $\frac{\pi}{4}$ is a constant, its derivative is $0$. (In the above $y= \cos x$ and $|y| \leq 1$ so we can use the arctangent addition formula). Nov 3 '18 at 2:47
• @YuriyS WOW, thanks. Nov 3 '18 at 2:51

## 1 Answer

There are two ways to show that both are anti-derivatives of he same functions.

The first way is to differentiate both of them and see if you get the same result.

The second is to show that the difference of you two solutions is a constant.

Each way has its own merit and sometimes one is easier to check than the other.

I prefer the first method because I also make sure that my answer is correct by comparing the derivatives with the original integrand for accuracy.

• nice answer, thanks. Nov 3 '18 at 4:15