@D_H comment #9,10:
f = sinx sec^n-1 x
f = tanx sec^n-2 x
f' = (n-2)tan^2 x sec^n-2 x + sec^n x
f' = ((n-2)sin^2 x +1)sec^n x
whats wrong -.-
@Count_Iblis comment#11 :
i've tried to think in another way...
i supposed that x^4/(x^2-1)^4 = A/(x-1)^4 + B/(x+1)^4
i found A = B = 1/16...
@Count_Iblis:
well how can integrate (1/(2(x-1))+1/(2(x+1)))^4?
@D_H
what do you exactly mean by "differentiate" ..?
i know that sinx sec^n-1 x is tan(x) * sec^n-2(x)
so the integration of tan(x) * sec^n-2 (x) is simply (sec^(n-1) (x))/(n-1) ... is that what i should find?
its not really a homework.. a friend of mine got challenged by this question by another friend by his tutor (as he said) so he came asking me how to solve it (as the class math guy).. anyway, in addition for further help, where can i find the... proof that the reduction formula is correct..
im...
Hello,
Recently, a friend challenged me to solve the integration of this function:
f(x) = tan^4(x)/cos^3(x)
After about a page of work, i found that i must integrate this:
x^4 / sqrt(x^2 -1)
Simply, the question is:
How can i integrate x^4/sqrt(x^2-1)?
I guess the answer is ln(2) * 2^x
No wonder i couldn't solve it before, they thaught us nothing about e
Depending on what you've said this means the derivative of e^x is e^x.. right?
Hi, I've been trying to find the derivative of 2^x and i got stuck here:
Lim (2^x - 1)/x
x=>0
How can i solve this limit (with steps please)
ps. I am only 17