evaluate indefinite integrals intergral tan^2x Sec^2x dx

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evaluate indefinite integrals intergral tan^2x Sec^2x dx

Calculus1
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tanx=u
take the derivative
derivative of tanx?

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Other answers:

both sides
what is it
derivative of tna is sec^2x
sec^2xdx=du
subs. them in int.
if tanx=u then what is tan^2x
u sqre
good
then what is your new int. depends on only u
int. tan^2x Sec^2x dx sec^2xdx=du tan^2x=u^2
csc x^3/ 3 ?
|dw:1355646302656:dw|
what are they in u
u^2 . du
yesssssss..
solve that integral..
1/3 tan^3 x+c ?
you got it..
but i have one question though
shoot it..
|dw:1355639354738:dw|
|dw:1355646949637:dw|
did not get it,s orry
they are the same thing but just different letter..
|dw:1355647192985:dw|
oh okay.
dx or du they are just saying that you are taking int. depends on that variable
|dw:1355647315134:dw|
you can assume that x is just a constant in here..
okay,thanks. i have another problem, will you help me with that?
ok..
thanks
|dw:1355640103615:dw|
|dw:1355647649309:dw|
take derivative
how do you decide which one is "U"?
(:
because I can see that derivative sinx =cosxdx and it is above
|dw:1355647787810:dw|
derivative of sin theta = cos theta d theta
but it is sinsqrt theta
cos sqrt theta d theta
|dw:1355647993776:dw|
one more thing youre missing
what is it?
derivative of sqrt theta
|dw:1355640793032:dw|
|dw:1355648238516:dw|
|dw:1355648283456:dw|
how did you get that half?
|dw:1355648328997:dw|
|dw:1355648354029:dw|
|dw:1355648407558:dw|
what is your new integral..
|dw:1355648571843:dw|
1/2 integral 1/sin sqrt theta +c?
|dw:1355648717143:dw|
|dw:1355648783325:dw|
confused
what is int. 1/x dx=?
d
x
|dw:1355648969297:dw|
okay
in your case lnu+c put u back in
I gotta go, any question..
so,thats the ans ? after putting the u?
yes, ln(sin sqrt theta)+c is the answer..
okay,thanks a lot
np..

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