Trignometric Integrals! help please!

[Guest]

Evaluate the trigonometric integral of:

/((secx)(tanx - secx))dx

and


/(tan^2(y)+1)dy

I'm not sure what I should substitute for "u"...please help me along the process

 

[galactus]

Evaluate the trigonometric integral of:

$\displaystyle \int\left[(secx)(tanx - secx)\right]dx$

$\displaystyle \int sec(x)tan(x)dx-\int sec^{2}(x)dx$

First part: $\displaystyle \int sec(x)tan(x)dx$

Let $\displaystyle u=sec(x), \;\ du=sec(x)tan(x)dx$

$\displaystyle \int du=u$

resub:

$\displaystyle =\boxed{sec(x)}$

Second part:

$\displaystyle sec^{2}(x)dx$

Let $\displaystyle u=tan(x), \;\ du=sec^{2}(x)dx$

$\displaystyle \int du=u$

$\displaystyle =\boxed{tan(x)}$

So, put it together and we have:

$\displaystyle \boxed{sec(x)-tan(x)}$

$\displaystyle \int (tan^{2}(y)+1)dy$

Use the identity $\displaystyle tan^{2}(y)=sec^{2}(y)-1$

Then, you can use the result from the first part.