# Thread: a very easy questin -integration

1. ## a very easy questin -integration

how to do this piece?

2. Originally Posted by afeasfaerw23231233
how to do this piece?
integrate by parts

let $\displaystyle u = \tan^2 x$ and $\displaystyle v' = 1~dx$

and recall: $\displaystyle \int uv' = uv - \int u'v$

3. The Pythagorean trig identity:
sin^2(A) +cos^2(A) = 1
Divide both sides by cos^2(A),
tan^2(A) +1 = sec^2(A)
tan^2(A) = sec^2(A) -1 --------------(i)

So,
INT.[tan^2(x)]dx
= INT.[sec^2(x) -1]dx
= tan(x) -x +C -------------------answer.

4. Originally Posted by ticbol
The Pythagorean trig identity:
sin^2(A) +cos^2(A) = 1
Divide both sides by cos^2(A),
tan^2(A) +1 = sec^2(A)
tan^2(A) = sec^2(A) -1 --------------(i)

So,
INT.[tan^2(x)]dx
= INT.[sec^2(x) -1]dx
= tan(x) -x +C -------------------answer.
very nice! and a lot easier than doing it by parts

5. Originally Posted by afeasfaerw23231233
how to do this piece?
I would suggest:

$\displaystyle \int \tan^2(x)~dx=\int \frac{\sin^2(x)}{\cos^2(x)}~dx=\int \sin(x)~\frac{\sin(x)}{\cos^2(x)}~dx$

Then try integration by parts

RonL

6. thanks