# Integration Coursework

• Nov 13th 2008, 10:32 AM
smwatson
Integration Coursework
Find

$\displaystyle \int (1+tan(x))/(1+cos(x)) dx$

and

$\displaystyle \int (1+tanh(x))/(1+cosh(x)) dx$
• Nov 13th 2008, 11:33 AM
Math_Helper
• Nov 13th 2008, 11:43 AM
Moo
Hello,

It is pretty straightforward if you use Weierstrass substitution :

$\displaystyle t=\tan^2 \frac x2$

$\displaystyle \cos(x)=\frac{1-t^2}{1+t^2}$

$\displaystyle \sin(x)=\frac{2t}{1+t^2}$

$\displaystyle \tan(x)=\frac{2t}{1-t^2}$

and $\displaystyle dx=\frac{2}{1+t^2} ~ dt$
• Nov 13th 2008, 01:22 PM
Mathstud28
Quote:

Originally Posted by smwatson
Find

$\displaystyle \int (1+tan(x))/(1+cos(x)) dx$

$\displaystyle \int\frac{1+\tan(x)}{1+\cos(x)}dx=\int\bigg[\frac{1}{\cos(x)+1}-\frac{\sin(x)}{\cos(x)(\cos(x)+1)}\bigg]dx$
• Nov 13th 2008, 01:36 PM
Krizalid
\displaystyle \begin{aligned} \int{\frac{1+\tan x}{1+\cos x}\,dx}&=\int{\frac{(1+\tan x)(1-\cos x)}{\sin ^{2}x}\,dx} \\ & =\int{(1+\tan x)\left( \csc ^{2}x-\cot x\csc x \right)\,dx} \\ & =\int{\csc ^{2}x\,dx}-\int{\cot x\csc x\,dx}+\int{\frac{dx}{\sin x\cos x}}-\int{\csc x\,dx}. \end{aligned}

Each one of these are pretty staightforward to make.