1. ## integration by subsitiution

secx^3(tan x) dx

I'm not sure if my work is correct, but I know the answer is right because I checked it in my book. Can someone check the work in blod type, please?

u = secx
du/dx = secxtanx
du = secxtanx dx
du/secx = tanx dx

u^3*du/secx

secx^3/secx du

secx^2 du

secx^3/3

1/3(secx^3) + C

2. Originally Posted by zachb
secx^3(tan x) dx

I'm not sure if my work is correct, but I know the answer is right because I checked it in my book. Can someone check the work in blod type, please?

u = secx
du/dx = secxtanx
du = secxtanx dx
du/secx = tanx dx

u^3*du/secx

secx^3/secx du

secx^2 du

secx^3/3

1/3(secx^3) + C
How can you integrate sec(x^2) with respect to du??

I presume this is supposed to be:
Int[sec^3(x) * tan(x) dx]

= Int[sin(x)/cos^4(x) dx]

Let u = cos(x) ==> du = -sin(x) dx

Int[sec^3(x) * tan(x) dx] = Int[sin(x)/cos^4(x) dx] = Int[1/u^4 * (-du)]

= -Int[1/u^4 du] = -(-1/(3u^3)) + C

= 1/(3*cos^3(x)) + C = (1/3)*sec^3(x) + C

You apparently have the correct answer, but it looks like you had the wrong way to get it.

-Dan

3. Int[sin(x)/cos^4(x) dx]
I actually thought of writing it that way while I was away from my computer.
-Int[1/u^4 du] = -(-1/(3u^3)) + C
I would have done it like this:

-Int[1/u^4 du] = -Int[u^-4 du] = -(-1/3u^-3)) +C

But I'm assuming you left it the other way so that when you put cosx back into the function it would become secx.

Thanks.

4. Originally Posted by zachb
I would have done it like this:

-Int[1/u^4 du] = -Int[u^-4 du] = -(-1/3u^-3)) +C

But I'm assuming you left it the other way so that when you put cosx back into the function it would become secx.

Thanks.
Actually no. How you do the integral is a matter of taste. It's just the way I think of how to do the integral is all.

-Dan