What is #int_(0)^(1) (e^(2x) - e^(-2x)) / (e^(2x) + e^(-2x))dx #?

2 Answers
Sahar Mulla ❤ · Rhys · Mahek ☮
Apr 9, 2018

Let #(e^(2x) + e^(-2x)) = t#

# dt = [2e^(2x) -2e^(2x)]dx #

#dt/2=[e^(2x)-e^(-2x)]dx#

The given integral is, #int_0^1 (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx#

#=>int 1/t dt/2 #

#=>1/2int dt/t #

#=>1/2[logt] #

#=>1/2 [log(e^(2x) + e^(-2x))]_0^1 #

#=>1/2 [log (e^2+1/(e^2))] - 1/2 log 2 #

#1/2 [log (e^2/2+1/(2e^2))] #

Steve M
Apr 9, 2018

# int_0^1 \ (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)) \ dx =1/2 \ ln ((e^2+1/e^2)/2) ~~ 0.66250 #

Explanation:

We seek

# I=int_0^1 \ (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)) \ dx #

Noting that:

# sinh(x)=(e^(x)-e^(-x))/2# and # cosh(x)=(e^(x)+e^(-x))/2#

Then we can wrote:

# I=int_0^1 \ sinh(2x)/cosh(2x) \ dx #

# \ \ =int_0^1 \ tanh(2x) \ dx #

# \ \ =[1/2ln(cosh2x)]_0^1 #

# \ \ =1/2{ln(cosh2)-ln(cosh0)) #

# \ \ =1/2{ln((e^2+1/e^2)/2)-ln((1+1)/2)} #

# \ \ =1/2 \ ln ((e^2+1/e^2)/2) #

# \ \ ~~ 0.66250 #

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