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### Integration

```
Date: 02/17/99 at 17:48:47
From: crys
Subject: Integration

Can you help me with

(integral sign) x tan^2 x dx   ?

I think you have to use some kind of substitution or maybe trigonometry
identities

Thank you.
```

```
Date: 02/18/99 at 01:28:50
From: Doctor Luis
Subject: Re: Integration

The first thing that comes to mind is integration by parts. My primary
motivation for this method is the factor of x that appears on the
integrand, because I know that differentiating that factor will give
me a constant.

Quick review: integration by parts is essentially based on the
following formula

/                   /
|                   |
|  u dv  =  u*v  -  | v du
|                   |
/                   /

In this case the substitutions should be

u = x              ----->     du = dx

dv = tan^2(x) dx    ----->      v = ?

Note that there is a small problem here. We need to integrate the
function tan^2(x) in order to obtain v from dv. But this does not
really present that much of a problem, if you remember that
tan(x) = sin(x)/cos(x) and that cos^2(x)+sin^2(x) = 1.

/                /
|  sin^2(x)      |  1 - cos^2(x)
v =  | --------- dx = | ------------- dx
|  cos^2(x)      |    cos^2(x)
/                /

/
|
=  | (sec^2(x)-1)dx = tan(x)-x
|
/

Therefore, v = tan(x)-x. Now, substituting into our "integration by
parts formula," you get

/                                     /
|                                     |
| (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx)
|                                     |
/                                     /

where I have enclosed the substitutions we made in parentheses.

Now, the second term on the integral on the righthand side of the
equation presents no problem. It's just (1/2)x^2, but the first term
seems a bit more difficult. However, if you rewrite tan(x) as
sin(x)cos(x)/cos^2(x), you can see that we almost have a complete
differential; the derivative of cos^2(x) is 2*cos(x)*(-sin(x)).
So, concentrating on the integral of tan(x), we have:

/            /
|            |  sin(x)cos(x)
| tan(x)dx = | ------------- dx
|            |   cos^2(x)
/            /

If you make the substitution u = cos^2(x), which implies that
du = 2cos(x)*(-sin(x))dx or cos(x)sin(x)dx = -du/2), then this last
integral becomes

/                        /
|  sin(x)cos(x)       1  |  du
| ------------- =  - --- | ----- = -(1/2)ln(u) = -(1/2)ln(cos^2(x))
|    cos^2(x)         2  |   u
/                        /

Therefore the integral of tan(x) is just -(1/2)ln(cos^2(x)) and so our
original integral becomes:

/                                     /
|                                     |
| (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx)
|                                     |
/                                     /

= x*tan(x)-x^2 - (-(1/2)ln(cos^2(x))-(1/2)x^2)

= x*tan(x) - (1/2)x^2 + (1/2)ln(cos^2(x))

I hope this explanation helps, although it would have been better if
you had included an explanation of how you attempted this problem and
where you got stuck. (I did not know how or why you got stuck, so I
tried to explain everything.) Feel free to reply if you have any

Sincerely,

- Doctor Luis, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Trigonometry

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