Integration Coursework

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November 13th 2008, 10:32 AM
smwatson

Integration Coursework

Find

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

and

$\int (1+tanh(x))/(1+cosh(x)) dx$
November 13th 2008, 11:33 AM
Math_Helper

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Attachment 8680
November 13th 2008, 11:43 AM
Moo

Hello,

It is pretty straightforward if you use Weierstrass substitution :

$t=\tan^2 \frac x2$

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

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

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

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

Quote:

Originally Posted by smwatson

Find

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

$\int\frac{1+\tan(x)}{1+\cos(x)}dx=\int\bigg[\frac{1}{\cos(x)+1}-\frac{\sin(x)}{\cos(x)(\cos(x)+1)}\bigg]dx$
November 13th 2008, 01:36 PM
Krizalid

$\begin{aligned}<br /> \int{\frac{1+\tan x}{1+\cos x}\,dx}&=\int{\frac{(1+\tan x)(1-\cos x)}{\sin ^{2}x}\,dx} \\ <br /> & =\int{(1+\tan x)\left( \csc ^{2}x-\cot x\csc x \right)\,dx} \\ <br /> & =\int{\csc ^{2}x\,dx}-\int{\cot x\csc x\,dx}+\int{\frac{dx}{\sin x\cos x}}-\int{\csc x\,dx}.<br /> \end{aligned}$

Each one of these are pretty staightforward to make.

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