Another way to prove $\int_x^1 \frac{dt}{1+t^2}=\int_1^\frac{1}{x}\frac{dt}{1+t^2}$

Question

In order to prove $$ \int_x^1 \frac{dt}{1+t^2}=\int_1^\frac{1}{x}\frac{dt}{1+t^2}$$ I showed that $$ \int_x^1 \frac{dt}{1+t^2}- \int_1^\frac{1}{x} \frac{dt}{1+t^2}=\frac{\pi}{4}-arctan(x)-[arctan\bigg(\frac{1}{x}\bigg)-\frac{\pi}{4}]$$

Since $arctan(x)+arctan(\frac{1}{x})=\frac{\pi}{2}$, then it follows that $ \int_x^1 \frac{dt}{1+t^2}- \int_1^\frac{1}{x} \frac{dt}{1+t^2}=0$. However, I feel this is kind of proof only "shows" but doesn't really "explains".

Is there another way to prove this? The simpler the better!

Thanks in advance.

Answer 1

Let $I=\int_x^1 \frac{dt}{1+t^2}$. Make the substitution $t\to \frac{1}{t}$ and we get

$$I= \int_{\frac{1}{x}}^1 \frac{d(1/t)}{1+(1/t)^2}$$ $$=- \int_{\frac{1}{x}}^1 \frac{dt}{t^2(1+(1/t)^2)}$$ $$=- \int_{\frac{1}{x}}^1 \frac{dt}{(t^2+1)}$$ $$=\int_1^{\frac{1}{x}} \frac{dt}{(t^2+1)}$$

Answer 2

Hint: use the substitution $t=\frac{1}{x}$.

Answer 3

Let $l(x), r(x)$ be the left and right hand sides respectively.

Note that $l(1) = r(1) = 0$.

Note that $l'(x) = -{ 1\over 1+x^2}$, $r'(x) = { 1\over 1+x^2} {- 1 \over {1 \over x^2}}$. Since $l'(x) = r'(x)$ we have $l(x) = r(x)$ for $x >0$.