d/dx[int_1^tanx sqrt(tan^-1t)dt]?
2 Answers
By the FTC, Part 1:
And by the chain rule:
So:
d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = sec^2xsqrt(x)
Explanation:
If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.
The FTOC tells us that:
d/dx \ int_a^x \ f(t) \ dt = f(x) for any constanta
(ie the derivative of an integral gives us the original function back).
We are asked to find:
d/dx \ int_1^tanx sqrt(tan^-1t) \ dt
(notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly).
We can manipulate the definite integral using a substitution and the chain rule. Let:
u=tanx => (du)/dx = sec^2x
The substituting into the integral we get:
d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = d/dx \ int_1^u sqrt(tan^-1t) \ dt
" " = (du)/dx * d/(du) \ int_1^u sqrt(tan^-1t) \ dt
" " = sec^2x* d/(du) \ int_1^u sqrt(tan^-1t) \ dt
And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:
d/(du) \ int_1^u sqrt(tan^-1t) \ dt = sqrt(tan^-1u)
And so:
d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = sec^2x* sqrt(tan^-1u)
" " = sec^2x* sqrt(tan^-1(tanx))
" " = sec^2xsqrt(x)