Question

Integral of sin^3x/cosx dx

Answer 1

split up
\(\sin^3x dx= \sin^2x(\sin x dx)\)
and plug in u = cos x dx

du =... ?

Answer 2

du = -sin x dx

Answer 3

so now what does your integral change to ?

Answer 4

use sin^2 = 1-cos^2
first

Answer 5

\[\frac{ \sin^2x \sin x }{ \cos x } dx\]

then

\[\frac{ 1-\cos^2x \sin x}{ \cos x } dx ?\]

Answer 6

don't forget to use the brackets
\(\large \frac{( 1-\cos^2x )}{ \cos x }\sin x dx \)

Answer 7

and now just plug in cos x = u
sin x dx = -du

Answer 8

did u get something like
-(1-u^2)/u du
can u integrate this ?

Answer 9

\[\int\limits_{}^{}\frac{ 1-u^2 }{ u } du\]

Answer 10

with the negative sign

Answer 11

\[\int\limits_{}^{} u^{negative1} - u du\]

Answer 12

with a negative sign

Answer 13

\[-\ln u - \frac{ 1 }{ 2 }u^2\]

Answer 14

So the answer:

\[\ln cosx - \frac{ 1 }{ 2 }\cos^2x\]

Answer 15

-(1/u - u) du
-----> -ln u +u^2/2+c
just sign errors

Answer 16

So then \[-\ln cosx + \frac{ \cos^2x }{ 2 }\]

Answer 17

+c
yes, now correct :)

Answer 18

- ln cos x can also be written as ln sec x ,i f you want to.

Answer 19

Thank you! Does it need to be abs value?

Answer 20

since cos can take negative values, but natural log (ln) cannot, we need to specify, the absolute value , like \(\large \ln|\cos x|\)

Answer 21

or \(\large \ln |\sec x|\)

Answer 22

Thanks again!!! =)

Answer 23

welcome ^_^