Evalutating the Integral: int [a,b] [ (1-sinx) / cosx ] dx

[skatru]

I feel like I've evaluated this integral correctly but the answer in the book is different than what I am getting. Could you just inform me if I am right or if the book is right? Thanks.

\(\displaystyle \int_a^b \frac{1 - sin x}{cosx} dx\)

\(\displaystyle \int_a^b \frac{1}{cosx}(1 - sinx) dx\)

\(\displaystyle \int_a^b \frac{1}{cosx} - \frac{sinx}{cosx} dx\)

\(\displaystyle \int_a^b secx - tanx dx\)

= ln(sec x - tan x) - ln(sec x) + C

That's what I got and basically how I got the answer. The answer the book gives is

ln(1 + sinx x) + C

Am I missing something?

 

[pka]

Re: Evalutating the Integral

I feel like I've evaluated this integral correctly but the answer in the book is different than what I am getting. Could you just inform me if I am right or if the book is right? Thanks.

\(\displaystyle \int_a^b \frac{1 - sin x}{cosx} dx\)

\(\displaystyle \int_a^b \frac{1}{cosx}(1 - sinx) dx\)

\(\displaystyle \int_a^b \frac{1}{cosx} - \frac{sinx}{cosx} dx\)

\(\displaystyle \int_a^b secx - tanx dx\)

= ln(sec x - tan x) - ln(sec x) + C

That's what I got and basically how I got the answer. The answer the book gives is

ln(1 + sinx x) + C

Am I missing something?

You are both correct.
\(\displaystyle \ln (\sec (x) - \tan (x)) - \ln (\sec (x)) = \ln \left( {\frac{{\sec (x) - \tan (x)}}{{\sec (x)}}} \right) = \ln \left( {1 + \sin (x)} \right)\)

 

[skeeter]

Re: Evalutating the Integral

check it out ...

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

\(\displaystyle \int \frac{\cos{x}}{1+\sin{x}} \, dx = \ln(1 + \sin{x}) + C\)

 

[Deleted member 4993]

Re: Evalutating the Integral

I feel like I've evaluated this integral correctly but the answer in the book is different than what I am getting. Could you just inform me if I am right or if the book is right? Thanks.

\(\displaystyle \int_a^b \frac{1 - sin x}{cosx} dx\)

\(\displaystyle \int_a^b \frac{1}{cosx}(1 - sinx) dx\)

\(\displaystyle \int_a^b \frac{1}{cosx} - \frac{sinx}{cosx} dx\)

\(\displaystyle \int_a^b secx - tanx dx\)

= ln(sec x + tan x) - ln(sec x) + C<<<< Now it can convert to your book's answer.

That's what I got and basically how I got the answer. The answer the book gives is

ln(1 + sinx x) + C

Am I missing something?