Intergration and differentiation experts please help!

[Jones546]

The question is in the attachment...someone please help!

[System_Error]

Several ways of doing this. It's possible to use 'u' substitution, but I'm not sure it would work out that well in this case. Remember the chain rule (since there are multiple functions) and you should be fine.

[Rich2189]

Not 100% sure but,

1/an + a

1/3(cosx -sinx)^3

[zaza]

No no no. There are functions of x in the brackets, so you need to remember the chain rule. As said in post #2. If all else fails, just expand out the brackets and treat each of the three resultant terms (cos^2, cos sin and sin^2) as separate integrals.


zaza

[jemidiah]

It simplifies enough to be done in your head (even though I forgot about the +x at first, and mixed up my sin/cos ).

cos^2 (x) + x + C = y

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Expand the square, and you'll see what I mean:

(cosx-sinx)^2 = cos^2 (x) -2sinxcosx + sin^2(x) = 1 - 2sinxcos

It then is a very simple u/du substitution:

u = cosx
du = -sinxdx
int(-2sinxcosx dx) = int(2udu) = u^2 + C = cos^2 (x) + C

I neglected to include the 1 in the above integration, but the integral of 1 is x, and the integral of a sum is the sum of the integrals, so:

y = x + cos^2 (x) + C

Substitute your initial values and you're done.