Recent content by KDeep

Nvm.. its not the same thing.

u^5 - u^7 = -u^2?

Am I doing good on this problem?

u = x^2 du = 2xdx du/2 = xdx ∫xtan^1(x^2)dx (1/2)∫tan^-1(u)du I cannot figure out the integral of tan^-1(u)

u = x^2 du = 2xdx du/2 = xdx ∫xtan(x^2)dx (1/2)∫tan(u)du (1/2) ln |sec(x^2)| + C Correct?

Homework Statement ∫xtan^-1(x^2)dx Homework Equations The Attempt at a Solution I did u = x, du = 1, v = ? ,dv = tan^-1(x^2)dx I do not know how to get the integral of tan^-1(x^2)

I am confident that the derivative of ln| secθ + tanθ | is secθ.

I have never heard about hyperbolic sine function at all, I will definitely look into that. My Professor wants us to do trig substitutions since it is in the section.

Correct, so its 1/4∫ secθ dθ which is ln | secθ + tanθ | + C 1/4ln | (x+2/2) + (x/2) | is the answer right?

Homework Statement Making sure I got the right answer.Homework Equations The Attempt at a Solution

The integral is -cotθ + C

Yeah thanks for the tip. So the answer is 1/(9√3)? Am I correct?

Ok that's what I was going for. So tan(sin-1(x/3)) (sin-1((3/2)/3)) is ∏/6 right? 1/9 (Tan (∏/6) - Tan(0)) ? Am I on the right track?

A = 0 , B = 3/2 ∫ dx/((9-x2)3/2) First I substituted , x = 3sin∅ , dx = 3cos∅d∅ ∫ 3cos∅d∅/(32-(3sin∅)2)3/2 Then I factored. ∫ 3cos∅d∅/(32(1-sin2∅))3/2 Then I used the trig identity to convert 1-sin2∅ to cos2θ ∫ 3cos∅d∅/(32cos2∅)3/2 Then I simplified (1/9) ∫ (1/cos2θ) d∅ Then I...