evaluate intergral cosh(x)^(1/2)/(x)^(1/2)dx, etc

[abby_07]

can someone please help me on evaluating these intergrals.

1) cosh(x)^(1/2)/(x)^(1/2)

This is what I did, but I do not think it is right. First, I made two integrals so that I have:

cosh(x)^(1/2)/1 * 1/(x)^(1/2)

Then, using the formula d/dx[cosh u]=(sinh u)u', I got:

(x)^(1/2)(sinh(x)^(1/2))

Then I integrated 1/(x)^(1/2) and got 1/2(x)^(1/2), so I multiplied them and got:

(x)^(1/2)(sinh(x)^(1/2))/2(x)^(1/2)

2) Evaluate the integral sinh^2xdx

This looks like an identity but I do not know what to do. Can you please help?
__________________
Edited by stapel -- Reason for edit: punctuation, capitalization, spelling, etc

 

[skeeter]

Re: evaluate the intergral

can someone please help me on evaluating these intergrals.
cosh(x)^(1/2)/(x)^(1/2)
this is what i did but i do not think it is right

i made two intergrals so that i have
cosh(x)^(1/2)/1 * 1/(x)^(1/2)
then useing the formula d/dx[cosh u]=(sinh u)u'
i got
(x)^(1/2)(sinh(x)^(1/2))

then i intergrated 1/(x)^(1/2) and got 1/2(x)^(1/2)

so i multiplied them and got
(x)^(1/2)(sinh(x)^(1/2))/2(x)^(1/2)

do you know the method of substitution?
let u = x^1/2
du = 1/(2x^1/2) dx

INT cosh(x^1/2)/x^1/2 dx =

2*INT cosh(x^1/2)* 1/(2x^1/2) dx =

2*INT cosh(u) du

integrate and back substitute.




the next problem is to evaluate the intergral
sinh^2xdx

this looks like an idenity but however i do not kow what to do can you please help

yes, you need an identity ...
sinh²x = [cosh(2x) - 1]/2

and, btw ... it's spelled integral.

 

[abby_07]

what do you mean by back substitute?

 

[skeeter]

what do you mean by back substitute?

after you find the antiderivative in terms of u, back substitute x^1/2 for u.