Solving for I1/I2 using substitution

[cr7einstein]

Homework Statement

Hi all,

This problem has been troubling me for a while now; even though I have tried my best ( and filled up a rough notebook in the process). Consider $$I_1=\int_{0}^{1} \frac{tan^{-1}x}{x} dx$$$, and $$I_2=\int_{0}^{\pi/2} \frac{x}{sinx}dx$$. We are supposed to find $$\frac{I_1}{I_2}$$. The answer is $$1/2$$.

Homework Equations

The Attempt at a Solution

My try- To make the limits for both identical, I substituted $$x=sin\theta$$ in the first integral, and then tried to make use of the properties of definite integrals ( replacing $$f(\theta)$$ by $$f(\pi/2-\theta)$$ etc), but no real progress was made. I then tried $$x=arcsint$$ for the second one, but no result.

Now I really doubt if there is something wrong with the question itself, or am I just being really silly. Please help me here. Thanks in advance!

 

[member 587159]

I did not look to closely at this exercise, so I might be misleading you here.

But, I saw, that when you substitute ##x = \arcsin t## in ##I_2##, you get:

$$I_2 = \int\limits_0^1 \frac{arcsin t}{t\sqrt{1-t^2}}dt$$ and this integral has the same bounds as ##I_1##, so maybe you can use the linearity of the definite integral or something like that?

 

[Ray Vickson]

Homework Statement

Hi all,

This problem has been troubling me for a while now; even though I have tried my best ( and filled up a rough notebook in the process). Consider $$I_1=\int_{0}^{1} \frac{tan^{-1}x}{x} dx$$$, and $$I_2=\int_{0}^{\pi/2} \frac{x}{sinx}dx$$. We are supposed to find $$\frac{I_1}{I_2}$$. The answer is $$1/2$$.

Homework Equations

The Attempt at a Solution

My try- To make the limits for both identical, I substituted $$x=sin\theta$$ in the first integral, and then tried to make use of the properties of definite integrals ( replacing $$f(\theta)$$ by $$f(\pi/2-\theta)$$ etc), but no real progress was made. I then tried $$x=arcsint$$ for the second one, but no result.

Now I really doubt if there is something wrong with the question itself, or am I just being really silly. Please help me here. Thanks in advance!

Try the substitution ##\arctan(x)=y## in ##I_1##.