How to find the integral of the secant function

frozen7
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\int \sec hx
I solve it in this way:
\int \arccos hx
\int \ln (x^2 + \sqrt{x^2 -1}) dx

Then, I substitute u = \ln (x + \sqrt{x^2 + 1})
then I get
x\ln(x + \sqrt{x^2 + 1}) - \int x/\sqrt{x^2 + 1}dx

and then I substitute v = x^2 + 1

x\ln(x + \sqrt{x^2 + 1}) -1/2 \int v^(1/2) dv
and finally I get the answer,
x\ln(x + \sqrt{x^2 + 1}) -\sqrt{x^2 + 1} + C

Am I right?
 
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Do the Tanh(x/2) substitution.
 
tan (x/2) substitution? How?
 
Your second integral doesn't equal your first integral.
sech(x) is NOT equal to arcosh(x)!

sech(x) is the RECIPROCAL* of cosh(x), arcosh(x) is the (functional) INVERSE of cosh(x).




*Sometimes called the multiplicative inverse.
 
Do you know how to integrate sec(x)? Because a lot of times you can follow very similar procecures working with hyperbolic functions as with trigonometric ones. You just have to be careful about the signs of terms in certain identiies (eg, sin^2(x)+cos^2(x)=1 becomes cosh^2(x)-sinh^2(x)=1). Another approach would be to write sech out in terms of e^x and e^-x, and then use the substitution u=e^x.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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