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I have this question here which says:

Find $g''(\pi)$ where $g(y)=\int_{3}^{y^{2}} (\int_{0}^{\sin(x)} \sqrt{1+t^{2}} dt) dx$

Now I'm not really sure how to approach this. My intuition was to notice that the integral on the inside has a bound of $\sin(x)$ so I figured it must be a constant which I could pull out.

I could then use the fundamental theorem of calculus in order to take the derivative of the integral that's left over and then take the derivative again and solve that. I feel like that's wrong though since I have an integral left over.

My next idea was to just use the fundamental theorem of calculus normally so I would take the derivative of the outermost integral using the fundamental theorem of calculus.

I would then substitute a $y^2$ for the $t^2$ and multiply by the derivative of $t^2$ or just $2t$ and I would end up with:

$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dt)$

But then I'm not sure if the differential quantity changes, namely:

$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dt$

or

$\int_{0}^{\sin(x)} 2y\sqrt{1+y^{4}} dy$

I am not allowed to go beyond single variable calculus (so no double integrals allowed) nor am I allowed to use methods taught outside a standard introductory calculus. So I can't use trigonometric substitution, partial fractions, Integration by parts etc...

Any ideas about this?

SECOND EDIT:

So I did the question as follows.

Let $f(x)=\int_{0}^{\sin(x)} \sqrt{1+t^{2}} dt$

Thus we have, $$g(y)=\int_{3}^{y^{2}} f(x)$$

$$g(y)=F(y^2)-F(3)$$ $$g'(y)=2yf(y^2)-0$$ $$g'(y)=2yf(y^2)$$ $$g''(y)=2yf'(y^2)(2y)+2f(y^2)$$ $$g''(y)=4y^2f'(y^2)+2f(y^2)$$

So, $$f(y^2)=\int_{0}^{\sin(y^2)} \sqrt{1+t^{2}} dt$$

and,

$$f'(y^2)=\cos(y^2)\sqrt{1+\sin^{2}(y^2)}$$

Therefore,

$$g''(y)=4y^2f'(y^2)+2f(y^2)$$ $$g''(y)=4y^2\cos(y^2)\sqrt{1+\sin^{2}(y^2)}+2\int_{0}^{\sin(y^2)} \sqrt{1+t^{2}}$$ $$g''(\sqrt{\pi})=-4\pi$$

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2 Answers 2

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You can first integrate inner integral which has a known integration. After integrating ,putting limits it becomes $$\frac {\sin (x)}{2}\sqrt {1+\sin^2 (x)}+\frac {1}{2}\ln (\sin (x)+\sqrt {1+\sin^2 (x)}) $$ and then you have integral wrt x. And now use fundamental theorem of calculus .

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  • $\begingroup$ That is true but the integral of the inside part requires trigonometric substitution which isn't taught in this particular course and thus I am not allowed to use it. $\endgroup$ Nov 25, 2016 at 6:56
  • $\begingroup$ You can then use by part method to evalute the integral . Trigonometric substitution isnt a compulsion . I think you might have learnt that $\endgroup$ Nov 25, 2016 at 7:06
  • $\begingroup$ We don't cover beyond u-substitution here so I can't use parts either. Only u-substitution allowed. $\endgroup$ Nov 25, 2016 at 7:08
  • $\begingroup$ Never mind I got it on my own. $\endgroup$ Nov 25, 2016 at 7:57
  • $\begingroup$ See my edit. I believe that is correct. $\endgroup$ Nov 25, 2016 at 8:43
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You can treat it in this way.

$g^{'}(y)=2y\int_0^{siny^2}\sqrt(1+t^2)dt$

So you can further have

$g^{''}(y)=2\int_0^{siny^2}\sqrt(1+t^2)dt+4ysiny cosy\sqrt(1+sin^2y) $

Then you can have

$g^{''}(\pi)=2\int_0^{sin^2\pi}\sqrt{1+t^2}dt$

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  • $\begingroup$ I'm not sure if Fubini's theorem applies here. $\endgroup$ Nov 25, 2016 at 7:11
  • $\begingroup$ I did it another way. I have edited my post to reflect this. $\endgroup$ Nov 25, 2016 at 8:43

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