Calculate the integral ∫(tanx+cotx)(tanx/(1+cotx))^2dx.

[lfdahl]

Calculate the integral:

\[I = \int_{0}^{\frac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \frac{\tan x}{1 + \cot x} \right )^2dx.\]

A solution without the use of an online integral calculator is preferred.

 

[Albert1]

Calculate the integral:

\[I = \int_{0}^{\frac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \frac{\tan x}{1 + \cot x} \right )^2dx.\]

A solution without the use of an online integral calculator is preferred.

Spoiler

$-2+3 ln2 $

 

[lfdahl]

Spoiler

-2+3 ln2 (hope no miscalculation)

Spoiler

My mistake! Your result is correct!
Please show your integration steps. Thankyou.

 

[Albert1]

Spoiler

My mistake! Your result is correct!
Please show your integration steps. Thankyou.

my solution:

Spoiler

with transformation $y=tan(x),dy=sec^2(x)dx$
$$I = \int_{0}^{\dfrac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \dfrac{\tan x}{1 + \cot x} \right )^2dx.$$
we have:
$$I=\int_{0}^{1}(\dfrac{1+y^2}{y})\times\dfrac {y^4}{(1+y)^2}\times \dfrac{1}{1+y^2}dy=\int_{0}^{1}\dfrac{y^3}{(1+y)^2}dy$$
$=(\dfrac{y^2}{2}-2y+3ln(1+y)+\dfrac{1}{1+y})\big|_{0}^{1}=-2+3 ln2$

 

[lfdahl]

my solution:

Spoiler

with transformation $y=tan(x),dy=sec^2(x)dx$
$$I = \int_{0}^{\dfrac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \dfrac{\tan x}{1 + \cot x} \right )^2dx.$$
we have:
$$I=\int_{0}^{1}(\dfrac{1+y^2}{y})\times\dfrac {y^4}{(1+y)^2}\times \dfrac{1}{1+y^2}dy=\int_{0}^{1}\dfrac{y^3}{(1+y)^2}dy$$
$=(\dfrac{y^2}{2}-2y+3ln(1+y)+\dfrac{1}{1+y})\big|_{0}^{1}=-2+3 ln2$

Thanks, Albert! - for your participation - and a nice solution.