Tan is just sin/cos so your expression becomes (-4sin^3(x)/cos(x))
which integrates to 4 ln(cos(x))-cos(2x)+c
that's all. No restrictions to x. Maybe wolfram gives a plot or something, and of course you cannot plot a graph for x being infinite
Hi all,
I did this problem for my assignment:
integrate: -4cos^2(x)tan^3(x)
I answered:
4ln(abs(cos(x))) - 2cos^2(x) + C
Would this answer be considered correct? Wolfram alpha has one extra step after this answer, something about restricting the x values? Thanks for any clarification.
Oh I think I understand. The thing is, Wolfram's second last step WAS the answer I posted, and they got from that answer to the one you gave. I was confused about how to go from mine to yours, but I think it's from using the (1/2)(1+cos(2x)) = cos^2(x) identity? Because the 1 ends up being absorbed into the C constant. Is that the step?
This is what wolfram says:
"Which is equivalent for restricted x values to:
= 4 log(cos(x))-cos(2 x)+constant"
where log is the natural log