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I am giving as homework to solve these integral

1) $\int \limits_{0}^{1} \frac{\arctan(y x)}{x \sqrt{1-x^2}}dx$

2)$\int \limits_{0}^{\frac{\pi}{2}} \frac{x}{\tan x}dx $ with the hint of $0\leq y \leq 1$ and $\int \limits_{0}^{\frac{\pi}{2}} \frac{\arctan( y \tan x)}{\tan x}dx $

If any one could solve any one of them because i am new to this and don't have a clue on how to proceed.

Thanks.

Ahmad
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    You can find here some examples: https://math.stackexchange.com/q/2987994/515527 that should help you later solve those and for the second one: https://math.stackexchange.com/q/2966938/515527 – Zacky Mar 05 '19 at 12:37

1 Answers1

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$$I_1(y)=\int_0^1 \frac{\arctan(yx)}{x\sqrt{1-x^2}}dx\Rightarrow I_1(0)=0$$ $$I_2(y)=\int_0^{\pi/2}\frac{\arctan(y\tan x)}{\tan x}dx\Rightarrow I_2(0)=0$$

First for $I_1$, take $\frac{d}{dy}$ on both sides (and use the Leibniz integral rule) $$I_1'(y)=\int_0^1\frac1{x\sqrt{1-x^2}}\frac{\partial}{\partial y}\arctan(yx)dx$$ $$I_1'(y)=\int_0^1\frac1{x\sqrt{1-x^2}}\frac{x}{1+y^2x^2}dx$$ $$I_1'(y)=\int_0^1\frac{dx}{(1+y^2x^2)\sqrt{1-x^2}}$$ Then set $x=\sin(u)$: $$I_1'(y)=\int_0^{\pi/2}\frac{du}{1+y^2\sin^2u}$$ Then set $u=t/2$: $$I_1'(y)=\int_0^\pi\frac{dt}{2+y^2-y^2\cos t}$$ Then let $x=\tan(t/2)$: $$I_1'(y)=2\int_0^\infty \frac{1}{2+y^2+y^2\frac{x^2-1}{x^2+1}}\frac{dx}{x^2+1}$$ $$I_1'(y)=\int_0^\infty \frac{dx}{(1+y^2)x^2+1}$$ And since it is easily shown that $$\int_0^\infty \frac{dx}{ax^2+1}=\int_0^\infty \frac{dx}{x^2+a}=\frac\pi{2\sqrt{a}}$$ We have that $$I_1'(y)=\frac\pi{2\sqrt{1+y^2}}$$ And since $I_1(0)=0$ we have that $$I_1(y)=\frac\pi2\int_0^y \frac{da}{\sqrt{1+a^2}}=\frac\pi2\sinh^{-1}(y)$$

Like with $I_1$, we compute $I_2$ by taking $d/dy$ on both sides which gives $$I_2'(y)=\int_0^{\pi/2} \frac1{\tan x}\frac{\partial}{\partial y}\arctan(y\tan x)dx$$ $$I_2'(y)=\int_0^{\pi/2} \frac1{\tan x}\frac{\tan x}{1+y^2\tan^2x}dx$$ $$I_2'(y)=\int_0^{\pi/2}\frac{dx}{1+y^2\tan^2x}$$ We can re write this as $$I_2'(y)=\int_0^{\pi/2}\frac{\sec(x)^2dx}{(1+\tan(x)^2)(1+y^2\tan(x)^2)}$$ We then use $u=\tan(x)$ to get $$I_2'(y)=\int_0^\infty \frac{du}{(1+u^2)(1+y^2u^2)}$$ Then we can use partial fractions and a trig sub to get $$I_2'(y)=\frac\pi{2y+2}$$ And since $I_2(0)=0$, we have that $$I_2(y)=\frac\pi2\int_0^y \frac{dt}{t+1}=\frac\pi2\ln(y+1)$$ So we plug in $y=1$ to get that $$\int_0^{\pi/2}\frac{x}{\tan x}=\frac\pi2\ln2$$

clathratus
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