Question

Integrate :-

Answer 1

@Fifciol

Answer 2

Break the

\[\tan ^3(x) = \tan^2(x)\tan(x)\]

then integrate by parts. See if that helps

Answer 3

∫tan³x dx= ∫tan²x tanx dx
=∫(sec²x−1)tanx dx
=∫sec²xtanx dx − ∫tanx dx
Put u=tanx, then du/dx=sec²x, du=sec²xdx
=∫u du − ∫tanx dx
=u²/2 − log|secx| + C
=(tan²x)/2 − log|secx| + C from yahoo answers

Answer 4

or use UV method

Answer 5

Ok...

Answer 6

\[\int\limits \tan^3xdx=\int\limits \tan^2*\tan xdx=\int\limits \frac{ \sin^2x }{ \cos^2x }\tan xdx=\int\limits \tan x \frac{ 1-\cos^2x }{ \cos^2x }dx\]=
\[\int\limits \tan x(\frac{ 1 }{ \cos^2x }-1)dx=\int\limits \frac{ \tan x dx }{ \cos^2 x }-\int\limits tanx dx\]

Answer 7

Taking U=(tanx)^3 and V=1!!

Answer 8

substitute u=tanx du=1/cos^2 x

Answer 9

so\[\int\limits udu=\frac{ u^2}{ 2 }+C=\frac{ \tan^2x }{ 2 }+C\]and\[\int\limits \tan xdx=\ln(cosx)+C\]

Answer 10

yw bro :)

Answer 11

Thank you .... I am a girl... Okay