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$$ \int_0^1 \! C(x) \, \mathrm{d} x. $$

where $C(x) = \int_0^x \cos(t^2) \, \mathrm{d} t$.

I am really not quite sure how to go about this one, especially given that it needs to be calculated using integration by parts.

My lecturer has an example (done using integration by parts) for $\int_0^1 \! xC(x) \, \mathrm{d} x$.

In this example, he let $u = C(x)$, so that $\frac{du}{dx}\ = \cos(x^2)$, which worked out nicely.

In the 2nd last line of his solution, he had the term: $$ \int_0^1 \! \sin(x^2) \, \mathrm{d} x. $$ and simply finished with leaving this part as $S(1)$.

However, in this question, I don't seem to have terms which I can choose as $u$ and $dv$?

Is anyone able to give me some direction?

Many thanks!

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up vote 1 down vote accepted

It's easier to do integration by parts here:

$$\int C(x)\mathrm dx=x\,C(x)-\int x\cos(x^2)\mathrm dx$$

Can you take it from here?

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Thanks J.M., just to clarify: you have simply chosen C(x) for u, and x (from dx) as v'? Then I could just apply integration by parts to the last part right? – JackReacher Aug 12 '12 at 10:13
Right you are. The last part is more easily done through substitution, though: $u=x^2$, $\mathrm du=2x\,\mathrm dx$... – J. M. Aug 12 '12 at 10:16
gotcha. Thanks! – JackReacher Aug 13 '12 at 4:11

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