Question

Integral of Sec^3x-secx dx
This is the last part of a much bigger problem (;

Answer 1

Integral (sec^3(x) dx)

First, split sec^3(x) into sec(x) and sec^2(x).

Integral (sec(x) sec^2(x) dx)

Now, use integration by parts.

Let u = sec(x). dv = sec^2(x) dx
du = sec(x)tan(x). v = tan(x)

Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec(x)tan^2(x) dx)

Use the identity tan^2(x) = sec^2(x) - 1.

Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec(x)[sec^2(x)
- 1] dx)

Distribute the sec(x).

Integral (sec^3(x) dx) = sec(x)tan(x) - Integral( (sec^3(x) - sec(x)) dx)

Now, separate into two integrals.

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

Distribute the minus over the brackets.

Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec^3(x) dx) +
Integral (sec(x) dx)

Here's the part which gets tricky; we're going to move
- Integral (sec^3(x) dx) to the left hand side of our equation, resulting in TWO of them.

2 Integral (sec^3(x) dx) = sec(x)tan(x) + Integral (sec(x) dx)

And we know what the integral of sec(x) is (we stated it above).

2 Integral (sec^3(x) dx) = sec(x)tan(x) + ln|sec(x) + tan(x)|

All we have to do now is divide everything by 2, which is the same as multiplying everything by (1/2).

Integral (sec^3(x) dx) = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|

And don't forget to add the constant.

= (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C

Answer 2

Im so sorry one last thing. It was supposed to be 9 times that integral. So on each piece I pull out of the integral will it also be multiplied by the constant 9? For example would it break into 9uv-int u'v?