Finding derivative of this integral function.

Question

I need help on finding the derivative of this:

$$g(x) = \int_1^{x^2} (x-t)\sin^2(t)dt$$

I thought about taking out x and having it as a constant but how?

Answer 1

Distribute the parentheses into two integrals and differentiate from there using the fundamental theorem.

$$g'(x) = \int_1^{x^2} \sin^2(t)dt + 2x^2\sin^2(x^2)-2x^3\sin^2(x^2)=\\ (-2 + 2 x^2 + \sin 2 - \sin(2 x^2))/4 + 2x^2\sin^2(x^2)-2x^3\sin^2(x^2)$$

Answer 2

From Fundamental theorem of calculus, we have:

$$g(x) = x\int_1^{x^2}\sin^2(t)dt-\int_1^{x^2} t\sin^2(t)dt=x[F(x^2)-F(1)]-[H(x^2)-H(1)]$$

where

$$F(x^2)=\int_1^{x^2}\sin^2(t)dt,\ \ \ H(x^2)=\int_1^{x^2}t\sin^2(t)dt$$

and from Product rule:

$$g'(x) =F(x^2)-F(1)+ 2x^2 F'(x^2)-2x H'(x^2)$$

To calculate it, you have to solve the integral

$$F(x^2)=\int_1^{x^2}\sin^2(t)dt$$