values of q for which tangent integral is converges

Question

Finding value of $q$ for which $$\int^{1}_{0}\frac{1}{(\tan (x))^{q}}dx$$ converges

What i try::

Let $\tan x=t.$ Then $\displaystyle dx=\frac{1}{\sec^2 (x)}dx=\frac{1}{1+t^2}dt$

And changing limits

$$I=\int^{\tan (1)}_{0}\frac{1}{(1+t^2)t^{q}}dt<\int^{\tan(1)}_{0}\frac{1}{t^2\cdot t^{q}}dt$$

$$I<\int^{\tan(1)}_{0}t^{-q-2}dt=\frac{1}{-q-1}\bigg(t^{-q-1}\bigg)\bigg|^{\tan(1)}_{0}=\frac{1}{-q-1}\cdot (\tan (1))^{-q-1}$$

Here $q\in\mathbb{R}-\{-1\}$ for which integral i converges

Can anyone plese explain me is my solution is right. If not then how do i solve it. Help me please.Thanks

Answer 1

The integral $\int_{0}^a \frac{dx}{x^p}$ is real and finite(converges) for $p <1$. As $\tan x \approx x$ the integral $I=\int_{0}^{1} \frac{dx}{\tan^q x}$ will converge when $q<1$.

Answer 2

As $x\to0$, $\tan x\sim x$ and so $I$ converges iff $\int_0^1\frac{dx}{x^q}$ converges, that is iff $q<1$.