Hyperbolic Integration: cos(x)/[1 + sin^2(x)]
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pka
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What are the exact instructions that go with this exercise?
As I said above, hyperbolic functions are not involved with ordinary trigonometric functions such as sine and cosine. Such functions form a class known as circular functions. Whereas, hyperbolic functions are based on the exponential function e<SUP>x</SUP>.
If you can give us the instructions, (the name of your textbook would be useful too), that go with the problem, then we may see what your author is trying to do.
As I said above, hyperbolic functions are not involved with ordinary trigonometric functions such as sine and cosine. Such functions form a class known as circular functions. Whereas, hyperbolic functions are based on the exponential function e<SUP>x</SUP>.
If you can give us the instructions, (the name of your textbook would be useful too), that go with the problem, then we may see what your author is trying to do.
The sensible way to do this problem is with a simple u substitution.
\(\displaystyle u=sin(x)\;\ du=cos(x)dx\)
This gives:
\(\displaystyle \L\\\int\frac{1}{1+u^{2}}du=tan^{-1}(u)=tan^{-1}(sin(x))\)
I've never seen hyperbolics involved in a problem like this. Interesting, yet nonsensical.
\(\displaystyle u=sin(x)\;\ du=cos(x)dx\)
This gives:
\(\displaystyle \L\\\int\frac{1}{1+u^{2}}du=tan^{-1}(u)=tan^{-1}(sin(x))\)
I've never seen hyperbolics involved in a problem like this. Interesting, yet nonsensical.
Hello, johnboy!
Are you sure there are no typos?
Nothing make any sense . . .
The only way to involve hyperbolic functions is to use the identities:
\(\displaystyle \;\;\sin x \:= \:-i\cdot\sinh(ix)\;\;\;\cos x\:=\:\cosh(ix)\)
and I don't think intend to involve complex variables.
If the original problem is correct: \(\displaystyle \L\,\int\frac{\cos x}{1\,+\,\sin^2x}\,dx\)
\(\displaystyle \;\;\)then let: \(\displaystyle u\,=\,\sin x\;\;\Rightarrow\;\;du\,=\,\cos x\,dx\)
Substitute: \(\displaystyle \L\,\int\frac{du}{1\,+\,u^2}\;\)\(\displaystyle =\;\arctan u \,+\,C\;=\;\arctan(\sin x)\,+\,C\)
If the original problem is: \(\displaystyle \L\,\int\frac{\cosh x}{1\,+\,\sinh^2x}\,dx\)
\(\displaystyle \;\;\)then let: \(\displaystyle u \,= \,\sinh x\;\;\Rightarrow\;\;du \,= \,\cosh x\,dx\)
Substitute: \(\displaystyle \L\,\int\frac{du}{1\,+\,u^2}\;\)\(\displaystyle =\;\arctan u\,+\,C\;=\;\arctan(\sinh x)\,+\,C\)
Are you sure there are no typos?
Nothing make any sense . . .
The only way to involve hyperbolic functions is to use the identities:
\(\displaystyle \;\;\sin x \:= \:-i\cdot\sinh(ix)\;\;\;\cos x\:=\:\cosh(ix)\)
and I don't think intend to involve complex variables.
If the original problem is correct: \(\displaystyle \L\,\int\frac{\cos x}{1\,+\,\sin^2x}\,dx\)
\(\displaystyle \;\;\)then let: \(\displaystyle u\,=\,\sin x\;\;\Rightarrow\;\;du\,=\,\cos x\,dx\)
Substitute: \(\displaystyle \L\,\int\frac{du}{1\,+\,u^2}\;\)\(\displaystyle =\;\arctan u \,+\,C\;=\;\arctan(\sin x)\,+\,C\)
If the original problem is: \(\displaystyle \L\,\int\frac{\cosh x}{1\,+\,\sinh^2x}\,dx\)
\(\displaystyle \;\;\)then let: \(\displaystyle u \,= \,\sinh x\;\;\Rightarrow\;\;du \,= \,\cosh x\,dx\)
Substitute: \(\displaystyle \L\,\int\frac{du}{1\,+\,u^2}\;\)\(\displaystyle =\;\arctan u\,+\,C\;=\;\arctan(\sinh x)\,+\,C\)