Question

integrate sex^2 xe^tanx dx from 0 to 1

Answer 1

\[\int\limits_{0}^{1}\sec ^{2}xe ^{tanx}dx\]

Answer 2

oops.... not "sex".... sec* ma bad...

Answer 3

substitue tanx=t
so u'll get sec^2xdx=dt and so u now need to integrate just e^t from 0 to pi/4.

Answer 4

x is in radians

Answer 5

int by parts im assuming

Answer 6

since sec^2 is the derivative of tan

Answer 7

yeah; it ints up to \(e^{tan(x)}\)

Answer 8

\[e^{tan(1)-e^{tan(0)}}\]
\[e^{tan(1)}-1\]

Answer 9

\[\int_{}Du.e^u.du \implies e^u\]
\[u = tan(x) \implies du = sec^2(x)\]
or some such logic right?

Answer 10

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Answer 11

lol..... sure

Answer 12

u can also do a simple subsitution where u = tanx , therefore du = sec^2 x
then plug in x=0 and x=1 to find the new values of u and integrate from 0 to pi/4 \[\int\limits_{0}^{\pi/4} e ^{u}du\]

Answer 13

^_^ Thanks a looot! :)

Answer 14

I'm not getting it quite straight with the different explanations but i think I understand the logic behind each method... thanks :)

Answer 15

yeah, the trick is to notice that the trig terms are really just derivatives of each other. this is really just asking you to find something like:

\(\int_{}2x\) \(e^{x^2}\) dx

Answer 16

this is a straight forward u - sub yes?
\[u=tan(x)\]
\[du=sec^2(x)dx\] but you should also change the limits of integration

Answer 17

yes what kanade said exactly

Answer 18

not even a u-sub; just a reintegration

Answer 19

sorry i didn't see it

Answer 20

thanks :) much appreciated