Question

Here's my first question: (FIrst part courtesy of Count Iblis)
Integral of x sqrt(19x-7)dx ?

Write the integral in terms of functins you do know the inegral of. Rewrite the factor of x as follows:

x = 1/19 (19 x) =

1/19 (19 x - 7 + 7) =

1/19 (19 x - 7) + 7/19

^ He wrote this down, but I don't know what to do with it. V_V

My next, and hopefully last question, is thus.
Integral of [(sin(7x)^2)*(sec(7x)^4) dt]

I flipped through my text book, and I can't find anything like it. According to my prof. I'm supposed to use Integration Tables.

Answer 1

<<Integral of x sqrt(19x-7)dx >>
Let u = 19x -7
dx = (1/19) du
x = (1/19)(u + 7)
So the integral becomes the sum of two integrals:

Integral of (1/19)u^(3/2) du
+ Integral of (7/19)u^(1/2) du

both of which are easy integrals. Remember to change from u back to (19x - 7) when you are done.

Answer 2

<<Integral of [(sin(7x)^2)*(sec(7x)^4) dt] >>

I assume your differential variable of integration is dx, not dt.

I suggest you rewrite sin(7x)^2 as 1 = cos^2(7x). That leaves you with two integrals: one involving sec^4(7x) and the other sec^2(7x) (since cos = 1/sec). Then substitute u for 7x and use a table of integrals. The integral of sec^2 du is tan u. There is a recursion formula for the integral of sec^n u du. It says that
Integral of sec^4 u du =
[sinu/(3 cos^3u)] + (2/3)tan u

Answer 3

Thank you so much. You cleared that up really well. Thank you.