Question

Find the indefinite integral. (Remember to use ln |u| where appropriate.)
∫e^-4x tan(e^-4x)dx

i really need help with this...the answer i'm getting is ln|cos e^-4x|

Answer 1

\[\Large \int\limits_{ }^{ }~e^{-4x}\tan(e^{-4x})~dx\]

Answer 2

Step 1:

First set u=-4x

(then du = -4 dx, and so you get,)

-1/4 du=dx


Step 2:
set w = e^u
dw = e^u du

then integrate the tangent....

Answer 3

\[\Large \int\limits_{ }^{ }~e^{-4x}\tan(e^{-4x})~dx \]\[u=-4x,~~-\frac{1}{4}du=dx\]\[\Large -\frac{1}{4}\int\limits_{ }^{ }~e^{u}\tan(e^{u})~dx\]

Answer 4

\[\Large -\frac{1}{4}\int\limits_{ }^{ }~e^{u}\tan(e^{u})~dx \]\[w=e^u,~~dw=e^u~du\]\[\Large -\frac{1}{4}\int\limits_{ }^{ }\tan(w)~dw\]

Answer 5

ohh, it is supposed to be du, where I have e^u, NOT dx..... sorry

Answer 6

let me know if you need help integrating the tangent

Answer 7

the tanget would be cosine right? @idku

Answer 8

not exactly.

Answer 9

If you don't remember the integral of tangent, then you can do this.
\[\int\limits_{ }^{ }\tan(w)dw=\int\limits_{ }^{ }\frac{\sin(w)}{\cos(w)}dw~~~~h=\cos(w),~-dh=\sin(w)\]\[-\int\limits_{ }^{ }1/h~dh=~-\ln(h)+C=-\ln(\cos~w)+C=\ln(\sec~w)+C\]

Answer 10

(I brought the exponent inside for the very las last step.

Answer 11

oh ok i see,
so it would be \[\ln|\sec e^-4x|+c\] ?

Answer 12

yes, ln ( sec e ^(-4x) ) + C

Answer 13

well, or,

- ln (cos e^(-4x) ) + C

Answer 14

ok, thanks

Answer 15

anytime