if $f(x) = \int_{t=1}^{t=x^2} t\sin^2(t)\operatorname d\!t$ then $\frac{\operatorname d\!f(x)}{\operatorname d\!x}=?$ $$f(x) = \int \limits_{t=1}^{t=x^2} t\sin^2(t)\operatorname d\!t$$
Do I use U-substitution and have the answer as $$f'(x) = 2x*x^2\sin^2(x^2)$$
Or does this question require integration by parts? 
Thanks.
 A: Your answer is correct. For example:
$$f(x)=\int_1^{x^2}t \sin^2(t)dt.$$
Using the substitution $t=x^2$ we have $dt=2xdx$. Hence,
$$f(t) = \int_1^t x^2\sin^2(x^2)2xdx = 2\int_1^t x^3\sin^2(x^2)dx.$$
Hence, we trivially have $$f'(t) = 2t^3\sin^2(t^2),$$ as required.
We can use Liebniz's more general rule of differentiation under the integral sign for more complicated integrals.
A: Why not to integrate directly? 
Think about this:


*

*Since $\sin^2 t = \frac{1-\cos{2t}}{t}$,

*Your integral becomes:


$$f(x) = \frac{1}{2}\int^{x^2}_{1} \left( t-t\cos{2t} \right) \, dt = \frac{1}{2} \left.\left( \frac{t^2}{2} - \frac{1}{4} \cos{2t} - \frac{1}{2} t \sin{2t} \right)\right|^{x^2}_1,$$
apply now Barrow's rule and differentiate after this.
Cheers!
A: Using integration by parts is how you would evaluate the integral however it is unnecessary since you are after the derivative of the integral.  From the fundamental theorem of calculus the derivative of an integral is merely the integrand function evaluated at the limit points.  We also must satisfy the chain rule.
So we simply plug in the limits of integration into the integrand and multiply each by the derivative of the respective limit.  The upper limit is a function so it remains while the lower limit is a constant making the derivative 0 so the whole term disappears.
$$
\left(x^2 \sin^2(x^2)\right)*2x - \left(1^2 \sin^2(1^2)\right)*0 = 2x*x^2 \sin^2(x^2)
$$
That matches your answer and you are done although completely simplified would be $2x^3 \sin^2(x^2)$.
