:( need more help, integration

[naz786]

Evaluate the integral: Integrate: (1-sinx)/ cosx

please help me it would be much appreciated

 

[Guest]

The best way I see to proceed is to seperate the fraction so you get:

\(\displaystyle [1/cosx] -[sinx/cosx]\)

Then if you integrate each bit seperately...

 

[Unco]

Following from O3jwood:

The second term (which is tan(x) but keep as sin(x)/cos(x)):
let \(\displaystyle \L u \, = \, \cos{x}\).

The first term (which is sec(x) and I've always disliked to integrate):

multiply sec(x) by \(\displaystyle \L \frac{\sec{x} + \tan{x}}{\sec{x} + \tan{x}}\)
and let \(\displaystyle u = \sec{x} + \tan{x}\).

I remember Metru (at SOS) using something more clever for the latter but I have forgotten.

 

[soroban]

Hello, naz786!

Want to see an off-the-wall method?

\(\displaystyle \L\int \frac{1\,-\,\sin x}{\cos x}\,dx\)

Multiply top and bottom by \(\displaystyle (1\,+\,\sin x):\)

. . \(\displaystyle \L\frac{1\,+\,\sin x}{1\,+\,\sin x}\cdot\frac{1\,-\,sin x}{\cos x}\;=\;\frac{1\,-\,\sin^2x}{\cos x(1\,+\,\sin x)} \;=\;\frac{\cos^2x}{\cos x(1\,+\,\sin x)} \;=\; \frac{\cos x}{1\,+\,\sin x}\)


We have: \(\displaystyle \L\,\int\frac{\cos x}{1\,+\,\sin x}\,dx\)

. . Let \(\displaystyle u\:=\:1\,+\,\sin x\;\;\Rightarrow\;\;du\:=\:\cos x\,dx\;\;\Rightarrow\;\;dx\:=\:\frac{du}{\cos x}\)

Substitute: \(\displaystyle \L\;\int\frac{\cos x}{u}\left(\frac{du}{\cos x}\right)\;=\;\int\frac{du}{u} \;= \;\ln|u| \,+\,C\)


Back-substitute: \(\displaystyle \L\;\ln|1\,+\,\sin x|\,+\,C\)

 

[royhaas]

Or multiply top and bottom by \(\displaystyle \cos x\) and use algebra and trig to end up with Soroban's result.