Posted by **Shelley** on Monday, January 29, 2007 at 1:23am.

Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you!

1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx

It says u = tan x to substitute

So if I use u = tan x, then my du = secx^2

then I have integral of (u^2)

then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?

__________________________________

2. integral from 0 to 2root3 of (x^3)/(root(16-x^2))

I am pretty sure this has something to do with trig substitution. is it something with tan-1 or cos(theta)?

I am not sure how to go about this one at all.

Thanks again! =)

. integral from 0 to pi/4 of (tanx^2)(secx^4)dx

It says u = tan x to substitute

So if I use u = tan x, then my du = secx^2

** no. du= sec^2x dx **
then I have integral of (u^2)

** Hmmm. INT u^2 sec^4x du/sec^2x
**

INT u^2 sec^2x du

how are you going to get rid of the sec^2 x?
second: draw the imaginary triangle, 4 is the hypotenuse, with angle theta , x is the opposite side, the adjacent side is sqrt (16-x^2)

then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?

## Answer this Question

## Related Questions

- Calc - Hello im trying to integrate tan^3 dx i have solved out the whole thing ...
- Calc - Hello im trying to integrate tan^3 dx i have solved out the whole thing ...
- Math - I can't find the integral for (tanx)^(6)*(secx)^(2) I tried splitting up ...
- calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...
- Calc - Evaluate the integral using any method: (Integral)sec^3x/tanx dx I ...
- Calculus - Int tanx sec^2x dx can be taken as (by putting it in form of Int xdx...
- Calculus - I think I have the correct answers for the following problems. For ...
- Trig - Verify that each of the following is an identity. tan^2x-sin^2x=tan^2xsin...
- integral Calculus - ʃ (4sin²x cos²×/sin2x cos2x)dx i've got it from the ...
- Math - Im really struggling with these proving identities problems can somebody ...

More Related Questions