Integral cos(x)/(1+sin^2(x)) dx

Messagehelp

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Sep 18, 2005
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I have to evaluate that integral. Im not sure about anything. the answer in the back says its tan^-1(sin(x))+c.

Thanks in advance
 
Hello, Messagehelp!

Please don't put the problem in the subject line.

\(\displaystyle \L\int\frac{\cos(x)}{1\,+\,\sin^2(x)}\,dx\)


Let u=sin(x)\displaystyle u\,=\,\sin(x) and substitute . . .
 
It should be du1+u2\displaystyle \frac{{du}}{{1 + u^2 }}.

That is arctangent form.
 
Hello, Messagehelp!

You already made an error . . .


\(\displaystyle \L\int\frac{\cos(x)}{1\,+\,\sin^2(x)}\,dx\)

Let u=sin(x)        du=cos(x)dx        dx=ducos(x)\displaystyle u\,=\,\sin(x)\;\;\Rightarrow\;\;du\,=\,\cos(x)\,dx\;\;\Rightarrow\;\;dx \,=\,\frac{du}{\cos(x)}

Substitute: \(\displaystyle \L\:\int\frac{\cos(x)}{1\,+\,u^2}\left(\frac{du}{\cos(x)}\right) \;=\;\int\frac{du}{1\,+\,u^2}\)

And you don't recognize this form??


Formula \(\displaystyle \L\;\int\frac{du}{1\,+\,u^2}\)  =  tan1(u)+C\displaystyle \;=\;\tan^{^{-1}}(u)\,+\,C
 
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