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# C4 Integration using trig and substitution help please. watch

1. ..
2. so you have cos^5 x
separate one cos away so you have cosx(cos^4 x)
rewrite cos^4 x as (1-sin^2 x)^2
so you have integral of cosx(1-sin^2 x)^2..

can you see where to go from there?
3. u cud de moivre dat ho, be like "bam bam *****eZ ima demoivre ur a\$\$\$\$" n den integr8 it dat way
4. (Original post by Farhan.Hanif93)
Would I then use a substitution of u=sinx?
try it.
5. (Original post by Farhan.Hanif93)
After using the substitution and integrating I now get:
(1 - sin^2x)^3
-------------------
6sinx

(the rather shoddy line represents a fraction. )
try this:

(integral of) cosx(1-sin^2 x)^2 dx
(rewriting it) = (1-sin^2 x)^2 cosx dx
u = sinx
du = cosx dx

Now you can replace the cosxdx in the integral with du

(1-(sin x)^2)^2 du

and because you made u = sinx, you can sub this in, and now you have everything in terms of u and du (which then you can integrate).
6. (Original post by Farhan.Hanif93)
Yes, this is what I did but I integrated (1-u^2)^2 using the reversed chain rule and the answer I got was as above.
Oh

the 'reserve chain rule' cant really be used here, you are assuming that if you differentiate (1-u^2)^3 / -6u you will get back to (1-u^2)^2 but you dont, because when you differentiate (1-u^2)^3 / -6u you have to use the chain rule AND the quotient rule giving you -36u^2(1-u^2)^2 - 6(1-u^2)^3 / 36u^2 which is quite messy.

Just expand (1-u^2)^2 and integrate
7. hello there, I manage to get this:

u-(2u^3)/3 +(u^5)/5 +c

I guess you then need to plug in u=sinx

thanks!

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