Calculating the derivative of an integral

Question

d/dx ₁ ∫^(1/x)  cos^3 (t) dt = (-1/x^2) cos^3(1/x) 

 

What I dont understand is where the -1/x^2 came from . I thought that the upper bound was just the derivative but isnt the derivative of 1/x ln abs x. 

Im so confused !

Answer 1

It comes from the chain rule.

 

If u is a function of x, a is a constant, and G(u)=∫_{(from t=a to t=u)}f(t)dt,

 

then (dG/dx) = (dG/du)(du/dx).

 

So, (d/dx)(∫_{(from t=1 to t = 1/x)}[cos³t]dt = [cos³(1/x)](1/x)'

 

                                                     = (-1/x²)cos³(1/x) 

Answer 2

This is how I would approach this problem.

 

Let  F(t) = ∫cos³tdt

 

Then the definite integral is           F(1/x) - F(1)

 

 

Rewrite the problem using trig identities.

 

F(t) = ∫cost(1 - sin²t)dt

F(t) = ∫(cost - costsin²t) dt

F(t) = sint - ∫(costsin²t)dt

 

Use substitution for the integral factor.

 

Let             u = sint

                du = costdt

 

∫sin²costdt = ∫u² du

                 = (1/3)u³

                 = (1/3)sin³t

 

Plugging back into the original integral.

 

F(t) = sint - (1/3)sin³t

 

Now evaluate     F(1/x) - F(1)