## anonymous one year ago where do I go wrong...

1. anonymous

|dw:1439117048654:dw|

2. anonymous

|dw:1439117111328:dw|

3. anonymous

4. ganeshie8

5. hartnn

your approach and answer is correct, you just got an equivalent answer! try to re-substitute u by sin 3t

6. hartnn

and then apply the double angle formula for cos 2x if you need the answer in terms of cos 6x

7. anonymous

in that case, finally! :D

8. hartnn

you got it ? you'd get sin^2 3t and would use $$\cos 2x= 1-2\sin^2 x$$ formula....

9. UsukiDoll

I don't see anything wrong with it either. the u-sub is correct u = sin3t du = cos3t(3) dx 1/3 du = cos3t dx

10. ganeshie8

you could also try |dw:1439117565392:dw|

11. hartnn

was just about to tell that^^ :P

12. UsukiDoll

then for double angle formula... notice that our original problem has a 3? $\cos 2x= 1-2\sin^2 x$ this is just standard... but by mutliplying 3 $\cos 2(3)x= 1-2\sin^2 3x$ $\cos 6x= 1-2\sin^2 3x$

13. UsukiDoll

multiply 3 on the left side for cos2(3) x ... but the 3 is needed to fit your question on the right hand side.

14. ganeshie8

remember, indefinite integral gives you a whole family of functions which are unique only upto a constant $$\cos 6x = 1-2\sin^23x$$ so, $$\cos 6x \equiv -2\sin^23x\pmod{\text{constant functions}}$$

15. anonymous

Really good tips guys, thank you all :)