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

[Messagehelp]

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

 

[soroban]

Hello, Messagehelp!

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

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


Let \(\displaystyle u\,=\,\sin(x)\) and substitute . . .

 

[Messagehelp]

i have worked it out to -du/1+u^2... but i dont know what to do next.

 

[pka]

It should be \(\displaystyle \frac{{du}}{{1 + u^2 }}\).

That is arctangent form.

 

[soroban]

Hello, Messagehelp!

You already made an error . . .


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

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

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