Question

Help please! anti derivative of sinx/cos^2/3 x

Answer 1

\[\Large\bf\sf \int\limits \frac{\sin x}{(\cos x)^{2/3}}dx\quad=\quad \int\limits \frac{1}{(\color{royalblue}{\cos x})^{2/3}}\left(\color{orangered}{\sin x\;dx}\right)\]

We need to make a u-sub,
\[\Large\bf\sf \color{royalblue}{u=\cos x}\]

Answer 2

What part are you stuck on? :)

Answer 3

I get stuck on the 2/3

Answer 4

Were you able to get your substitution setup correctly?
Do you know how to find your du?

Answer 5

du is -sinxdx

Answer 6

Ok good, moving the negative over,
\[\Large\bf\sf \color{orangered}{-du=\sin x\;dx}\]

Answer 7

\[\Large\bf\sf \int\limits\limits \frac{1}{(\color{royalblue}{\cos x})^{2/3}}\left(\color{orangered}{\sin x\;dx}\right)\quad=\quad \int\limits\limits \frac{1}{(\color{royalblue}{u})^{2/3}}\left(\color{orangered}{-du}\right)\]

Answer 8

So the 2/3 is messing you up?
Just bring it up to the numerator using rules of exponents.\[\Large\bf\sf \frac{1}{x^n}\quad=\quad x^{-n}\]From there you can simply apply the power rule!! :)

Answer 9

I get 3/5 u^5/3

Answer 10

Hmm did you remember that your u is negative? :O
I don't think you'll get all the way up to 5/3 power.
\[\Large\bf\sf \int\limits u^{-2/3}\;du\quad=\quad \frac{1}{-\frac{2}{3}+1}u^{-2/3+1}\quad=\quad \frac{1}{1/3}u^{1/3}\]

\[\Large\bf\sf =\quad 3u^{1/3}\]

Answer 11

Your exponent is negative* i meant :p

Answer 12

Duh! I forgot the negative! okay.. so is 3cos^1/3 x the antiderivative?

Answer 13

Mmm ya looks good!
With a +c somewhere at the end.

Answer 14

Thank You!