#d/dx[int_1^tanx sqrt(tan^-1t)dt]#?

2 Answers
Eddie
May 3, 2017

By the FTC, Part 1:

#d/dx (int_a^x f(t) dt) = f(x)#

And by the chain rule:

#d/dx (int_a^(g(x)) f(t) dt) = f(g(x)) (dg)/(dx)#

So:

#d/dx(int_1^tanx sqrt(arctan t ) \ dt)#

#= sqrt(arctan (tan x)) d/dx(tan x)#

#= sqrtx sec^2 x#

Steve M
May 3, 2017

# 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 constant #a#

(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)#

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