Why Is the Antiderivative of sec^2(4x+1)dx Not tan^4(4x+1)/4?

[SAT2400]

Homework Statement

1.∫sec^2 (4x+1)dx

2.∫ root(sin pi theta) cos(pi theta) d(theta)

3.∫ e^x dx /(1+e^x)

Homework Equations

1.u=4x+1
2. u= sin(pi theta)
3. u= 1+e^x
antiderivatives to find the answers..

The Attempt at a Solution

For #1, tan^4(4x+1)/4 +c..is the answer..but I don't get why tan should be ^4??
I took the deriv. of the u and replace dx with the stuff that only contains du..not dx...
And then,,I sometimes have problems solving some questions..
Please help!

 

[tiny-tim]

Hi SAT2400!

(have a pi: π and a theta: θ and a square-root: √ and try using the X² tag just above the Reply box )

1.∫sec^2 (4x+1)dx
…
For #1, tan^4(4x+1)/4 +c..is the answer..but I don't get why tan should be ^4??

It's a typo . tan(4x+1)/4 +c..is the correct answer!

2.∫ root(sin pi theta) cos(pi theta) d(theta)

3.∫ e^x dx /(1+e^x)
…
2. u= sin(pi theta)
3. u= 1+e^x
antiderivatives to find the answers..

(You mean ∫ √(sin(πθ)) cos(πθ) dθ ?)

Those substitutions should be ok (or you could just use u = e^x, with the same result) …

what did you get? ​