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$$F(t)=\int_{0}^{t^2}\mathbb{dx}\int_{x-t}^{x+t}\sin(x^2+y^2-t^2)\mathbb{dy}$$

find $F'(t)$

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Hint. $F(t)=\int_0^{t^2} g(x,t)\,dx$. Can you calculate $F'(t)$? –  vesszabo Nov 29 '12 at 17:46

1 Answer 1

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Recal that $$ \small F'(t)=\frac{d}{dt}\int\limits_{\alpha(t)}^{\beta(t)} f(t,x)dx= \int\limits_{\alpha(t)}^{\beta(t)} \frac{df}{dt}(t,x)dx+f(t,\beta(t))\frac{d\beta}{dt}(t)-f(t,\alpha(t))\frac{d\alpha}{dt}(t)\tag{1} $$ In your case $$ \small f(t,x)= \int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy, \qquad \alpha(t)=0,\qquad \beta(t)=t^2 $$ So after substitution we get $$ \small F'(t)=\int\limits_{0}^{t^2}dx\left( \frac{d}{dt}\int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy\right)+\left(\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy\right)\cdot 2t-\left(\int\limits_{0-t}^{0+t}\sin(0^2+y^2-t^2)dy\right)\cdot 0 $$ Thus $$ \small F'(t)=\int\limits_{0}^{t^2}dx\left( \frac{d}{dt}\int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy\right)+2t\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy \tag{2} $$ but this is not the end! Now you need to compute derivative of integral $$ \small \int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy $$ with respect to $t$. Though you have two parameters here you can think of $x$ as a number here. Formula $(1)$ will look like $$ \small G'(t)=\frac{d}{dt}\int\limits_{\gamma(t,x)}^{\delta(t,x)} g(t,x,y)dy =\int\limits_{\gamma(t,x)}^{\delta(t,x)} \frac{dg}{dt}(t,x,y)dy+g(t,x,\delta(t,x))\frac{d\delta}{dt}(t,x)-g(t,x,\gamma(t,x))\frac{d\gamma}{dt}(t,x)\tag{3} $$ In your case $$ \small g(t,x,y)=\sin(x^2+y^2-t^2)\qquad\gamma(t,x)=x-t\qquad\delta(t,x)=x+t $$ So after substitution we get $$ \small G'(t)=\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)(-2t)dy+\sin(x^2+(x+t)^2-t^2)\cdot 1-\sin(x^2+(x-t)^2-t^2)\cdot (-1) $$ $$ \small =-2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+\sin(2x^2+2xt)+\sin(2x^2-2xt) $$ $$ \small =-2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+2\sin(2x^2)\cos(2xt) $$ This finall result is $$ \small F'(t)=\int\limits_{0}^{t^2}dx\left( -2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+2\sin(2x^2)\cos(2xt)\right)+2t\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy $$

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