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:

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


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

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?


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




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


2 Answers 2


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 .

  • $\begingroup$ 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

You can treat it in this way.


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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.