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$$f(x) = \int \limits_{t=1}^{t=x^2} t\sin^2(t)\operatorname d\!t$$

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.

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  • $\begingroup$ You must integrate by parts. In order to make your life easier, uou could start replacing the square of the sine by something looking as the cosine of the double angle. $\endgroup$ – Claude Leibovici Feb 2 '14 at 8:59
  • $\begingroup$ Sorry ! I missed the fact that you wanted the derivative of $f(x)$ with respect to $x$. I thought you just wanted the integral. I apologize. By the way, your result is perfectly correct. $\endgroup$ – Claude Leibovici Feb 2 '14 at 9:10
  • $\begingroup$ @heropup it is the same person asking that question. I'm guessing he is using the answer here to answer the other one $\endgroup$ – Flowers Feb 2 '14 at 9:29
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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.

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  • $\begingroup$ So you mean: f'(x) = (2x^3)' * sin^2(x^2) + 2x^3(sin^2(x^2))'? $\endgroup$ – Kenneth Feb 2 '14 at 9:07
  • $\begingroup$ @Kenneth no I didn't mean that - see my revised answer. $\endgroup$ – Pixel Feb 2 '14 at 9:54
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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!

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  • $\begingroup$ Why would you integrate just to differentiate after? $\endgroup$ – Flowers Feb 2 '14 at 9:31
  • $\begingroup$ Yes, I know this is a bit too much waste of time. But, anyway, this intends to show that the integrand is easily integrable. Cheers! $\endgroup$ – Dmoreno Feb 2 '14 at 9:33
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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 $2x^3 \sin^2(x^2)$.

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