can anyone solve this calculus 2 problem Find 2x.tan^3.(x^2).(secx^2)to the power of-1/2dx?

Question

need step by step answer if possible

2x.tan^3.(x^2).sqrt(secx^2)dx

Answer 1

∫ 2x tan³(x²)/√sec(x²) dx

Substitute to get rid of the x² terms:

u = x²
du = 2xdx

Will change the problem to:

∫ tan³(u)/√sec(u) du

Then I think you can use the trig identity tan² = sec² - 1:

∫ tan(u)[sec²(u) - 1]/√sec(u) du

And split into two integrals:

∫ tan(u)sec²(u)/√sec(u) du - ∫ tan(u)/√sec(u) du

or, rewritten:

∫ tan(u)sec^(3/2)(u) du - ∫ tan(u)sec^(-1/2)(u) du

Now try another substitution:

v = sec(u)
dv = sec(u)tan(u) du

So split off a sec from each:

∫ sec^(1/2)(u) sec(u)tan(u) du - ∫ sec^(-3/2)(u) sec(u)tan(u) du

And substitute:

∫ v^(1/2) dv - ∫ v^(-3/2) dv

Now it's not too bad....

2/3 v^(3/2) + 2 v^(-1/2) + C

Substiting for v:

2/3 sec^(3/2)(u) + 2 sec^(-1/2)(u) + C

And then for u:

2/3 sec^(3/2)(x²) + 2 sec^(-1/2)(x²) + C

Answer 2

Please use parentheses and show us what part of that is raised to the power of -1/2.