Here it is
before we go on though, could you explain how we do this, since I tried a few methods, but all failed
Gladly ^^
It's been so long I can't find my original solution typed up in LaTeX ;_;
Anyway, utilizing the fact that 1/cos=sec, every term now has a sec factored in. So you have
Integral -1 to 1 of x2011*sec(x) + sec(x)tan(x) + sec(x)
Using even/odd identities, that an odd function times an even function results in an odd function, and seeing that x
2011 is odd while sec(x) is even, we know that their product is odd.
Thus the integral from -a to a of this product will always be zero.The second part simply evaluates to
sec(x)The third part is
ln(sec(x)+tan(x)) (Not gonna prove it here, just look up integral of sec(x))
And finally evaluating this result at the bounds, we get
sec(1) + ln(sec(1)+tan(1)) - sec(-1) - ln(sec(-1)+tan(-1))
Again, even/odd tells us for even (sec), f(-x)=f(x) and for odd(tan) f(-x)=-f(x)
So this is equal to
sec(1)+ln(sec(1)+tan(1)) - sec(1) - ln(sec(1)-tan(1))
Secants cancel and you're left with the logs. Using log rules, log(a) - log(b)= log(a/b), combine the two and you end up with
ln((sec2(1)-tan2(1))/(sec-tan)2)
Top goes to -1. Flip the 2 power to -2 and bring it up. Use log rules to bring down exponent as coefficient. Combine sec(1)-tan(1) using only cos and sin into -(cos(1))/(sin(1)-1). Negate the negative with the -1 we got earlier. Using the negative from the coefficient of the log, flip the inside fraction of the log (log rules), and you get Option B
I guess it was a little more work than I promised.. haha...
