that.
Answer:
f(u) = (u^(7/2) tan^14(sec^-1(√u)))/2
in the integral (tan^15(x)sec^8(x)dx
a substitution of u=sec^2(x) and the result is written as the integral f(u)du
find f(u)
I've gotten to the integral (tan^14(x)sec^6(x))/2 du but not sure where to go after
5 answers
I dont think thats the answer! would appreciate another look at it
AAAaannndd the bot gets it wrong yet again!
∫ tan^15(x) sec^8(x) dx = ∫ tan^15(x) sec^6(x) sec^2(x) dx
Let u = tanx, du = sec^2x dx. Then you have
∫ u^15 (u^2+1)^3 du = 1/22 u^22 + 3/20 u^20 + 1/6 u^18 + 1/16 u^16
now just reverse the substitution and you're done
∫ tan^15(x) sec^8(x) dx = ∫ tan^15(x) sec^6(x) sec^2(x) dx
Let u = tanx, du = sec^2x dx. Then you have
∫ u^15 (u^2+1)^3 du = 1/22 u^22 + 3/20 u^20 + 1/6 u^18 + 1/16 u^16
now just reverse the substitution and you're done
I tried entering the answer into my question but it still said it was wrong, also shouldnt u=sec^2(x) as per the question? I appreciate any help, been stuck on this for hours
I got the answer! for future readers, its ((u-1)^7u^3)/2