Recall that sin(x) = (e^ix-e^-ix)/(2i) and we can rewrite the integrand as {(e^ix-e^-ix)/(2i)}⁶Factor out {1/(2i)}⁶, which is -1/64, of the integrand (just to make algebra easier)Use binomial expansion to expand {(e^ix-e^-ix)}⁶ - with these types of questions you should find that the powers differ by multiples of 2 in the expansion (so we get 6,4,2,0,-2,-4,-6). You should get -(1/64)*{e^6ix - 6e^4ix + 15e^2ix - 20 + 15e^-2ix - 6e^-4ix + e^-6ix} = sin⁶(x) [remember the -64 which we factored out in step 3 to make the expansion less messy]Collect the terms which add 'nicely' to give something trigonometric. Remember that cos(nx) = (e^nix+e^-nix)/2 so you can rewrite sin⁶(x) as the sum of cosines.Then integrate this, remembering the factor of -1/64 and an arbitrary constant of integration (the '+c')If you feel you have time you can check with a calculator. Plug in some limits for the integration (maybe 1 and 2, for trig integrals avoid multiples of pi as they may be special cases) and use integration function on your calculator if it has it. Then recalculate the integral with your answer and they should be the same.
In summary for similar questions, remember to rewrite sin(x), or cos(x) in other examples, in terms of exponentials. Then expand and rearrange terms to then rewrite as the sum of cosines, or the sum of sines (often it is easier to factor out the power of 2 or 2i that appears). Then integrate term by term.

differentiate 2^x

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