Question

someone please help! ∫sin^3x / 1-cosx dx

Answer 1

Well, this is a nice substitution problem...
\[\Large \int \frac{\sin^3(x)}{1-\cos(x)}dx\]

Answer 2

Why don't you let u = cos(x) and work from there? :)

Answer 3

i think is better if u multiply with (1+cos(x))/(1+cos(x)) and see what u get :)

Answer 4

du^3 / 1- u???

Answer 5

whoa now... let's not get too far :P

Answer 6

Here, let me rearrange it for you..
\[\Large=-\int \frac{[\sin^2(x)][-\sin(x)]}{1-\cos(x)}dx\]

Answer 7

i tried multiplying both sides you get sin^3 x + sin^3x cosx all divided by sin 2 and then im stuck with a sinxcosx after simplifying it

Answer 8

So, letting u = cos(x)
du = -sin(x) dx

Answer 9

Work from there?

Answer 10

i thought the derivative of cosx is sinx ?

Answer 11

if you multiply with( 1+cosx)/(1+cosx) you get (sin^3x*(1+cosx))/(1-cos^2x)
then you have (sin^3(x)(1+cos(x)))/sin^2(x) , then its easy

Answer 12

No... the INTEGRAL of cos(x) is sin(x)
The derivative of cos(x) is -sin(x)

Answer 13

oh ok but if i sub it into the modified equation i get -du^2 +du / 1 - u

Answer 14

\[\int\limits\frac{ \sin ^{3} x }{1-\cos x }dx=\int\limits \frac{\sin ^{3}x }{ 1-\cos x }*\frac{ 1+\cos x }{ 1+\cos x }dx\]
\[=\int\limits \frac{ \sin ^{3}x \left( 1+\cos x \right) }{ 1-\cos ^{2}x }dx\]
\[=\int\limits \frac{ \sin ^{3}x \left( 1+\cos x \right) }{\sin ^{2}x}dx\]
\[=\int\limits \left( \sin x+\sin x \cos x \right)dx=-\cos x+\frac{ \sin ^{2}x }{ 2 }+c\]

Answer 15

@surjithayer thats so helpful but how did you get the integral of sinxcosx

Answer 16

\[\int\limits f ^{n }\left( x \right)f'\left( x \right)dx=\frac{ f ^{n+1}\left( x \right) }{ n+1 }\]