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### Integration by Parts

```
Date: 05/27/97 at 03:07:55
From: ajit
Subject: indefinite integrals

INT[sec^3 x]dx
```

```
Date: 05/27/97 at 10:33:44
From: Doctor Anthony
Subject: Re: indefinite integrals

If we factor sec^3 into sec^2(x)*sec(x), we can solve the integral
using parts.  Our integral becomes: INT[sec(x)*sec^2(x)dx].

INT[sec^3(x)dx] = I = sec(x)tan(x) - INT[tan(x)sec(x)tan(x)dx]

I  = sec(x)tan(x) - INT[sec(x)(sec^2(x)-1)dx]
I  = sec(x)tan(x) - INT[sec^3(x)dx] + INT[sec(x)dx]
2I = sec(x)tan(x) + INT[sec(x)dx]
2I = sec(x)tan(x) + INT[sec(x)(sec(x)+tan(x))/(sec(x)+tan(x)) dx]

Let u = sec(x)+tan(x), then du = (sec(x)tan(x) + sec^2(x))dx:

2I = sec(x)tan(x) + INT[du/u]
2I = sec(x)tan(x) + ln(sec(x)+tan(x)) + C
I = (1/2)sec(x)tan(x) + (1/2)ln[sec(x)+tan(x)] + C

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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