Question

how to integrate ((tan x)/(sec x + tan x)) dx

Answer 1

1/cos(x) + sin(x)/cos(x)

1+Sin(x)
---------
cos(x)

sin(x) cos(x)
----- * ------------
cos(x) 1+Sin(x)


sin(x)
-------
1+Sin(x)

Answer 2

i also get the same answer.. but i'm not sure whether this is the final answer??

Answer 3

\[\frac{\sin(x)}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)}=\frac{\sin(x)-\sin^2(x)}{1-\sin^2(x)}=\frac{\sin(x)-\sin^2(x)}{\cos^2(x)}=\frac{\sin(x)}{\cos^2(x)}-\frac{\sin^2(x)}{\cos^2(x)}\]
\[\frac{\sin(x)}{\cos^2(x)}-\tan^2(x)=\frac{\sin(x)}{\cos^2(x)}-(\sec^2(x)-1)=\frac{\sin(x)}{\cos^2(x)}-\sec^2(x)+1\]
just should really help you integrate the function

Answer 4

first one use a simple substitution
the other ones is just a quick memory away

Answer 5

can i simplify to be \[\sec ^{2} x ( \sin x - 1) + 1\] ??

Answer 6

\[\int sin(x)/ cos^2(x) dx\]

u= cos(x)

du=-sin(x) dx

\[\int -du/u^2\]

Answer 7

\[\int\limits_{}^{}\sin (x)/ \cos (x) +\int\limits_{}^{} \sec ^2 + \int\limits_{}^{} 1 dx\]
=1/cos (x) + tan (x) + x + C

Answer 8

\[\int\limits_{}^{}\frac{\sin(x)}{\cos^2(x)}dx-\int\limits_{}^{}\sec^2(x) dx+\int\limits_{}^{}1 dx\]
=1/cos (x) - tan (x) + x + C

Answer 9

forget the sign of \[\sec ^2 x\].. thanks a lot..

Answer 10

thats why i gave you a medal for having some correct things gj! :)

Answer 11

;)