2
$\begingroup$

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

Firstly I wanted to calculate $\int \sin (t^2) dt$ and then use $x$ and $\sqrt {x^2+1}$. But this antiderivative not exist so how can I do this? Is this function at all possible to count?

$\endgroup$
2
$\begingroup$

Let $f(t)=\sin(t^2)$.

$f$ is continuous at $\Bbb R$, thus

$\displaystyle{F: x\mapsto \int_0^xf(t)dt}$ is differentiable at $\Bbb R$ and for all $x\in \Bbb R$,

$$F'(x)=f(x)=\sin(x^2)$$

but

$$G(x)=\int_x^{\sqrt{x^2+1}}f(t)dt=$$ $$F(\sqrt{x^2+1})-F(x)$$

with $x\mapsto \sqrt{x^2+1}$ differentiable at $\Bbb R$. thus by chain rule, $G$ is differentiable at $\Bbb R$ and

$$G'(x)=\frac{2x}{2\sqrt{x^2+1}}F'(\sqrt{x^2+1})-F'(x)$$

$$=\frac{x}{\sqrt{x^2+1}}\sin(x^2+1)-\sin(x^2).$$

$\endgroup$
2
$\begingroup$

Hint: Use the Fundamental Theorem of Calculus, which says that if $$I(x) = \int_0^x \sin(t^2)\,{\rm d}t,$$then $I'(x) = \sin(x^2)$. Now observe that $f(x) = I(\sqrt{x^2+1}) - I(x)$ and use the chain rule.

$\endgroup$
0
$\begingroup$

Use the Lebinitz rule.

$$f'(x) = \sin(x^2+1) \times \frac{2x}{2\sqrt{x^2+1}} - \sin(x^2)$$

$\endgroup$
0
$\begingroup$

Let $g(x)=\int_0^{x} \sin (t^{2})dt$. Then $f(x)=g(\sqrt {1+x^{2}}) -g(x)$. ALso $g'(x)=\sin (x^{2})$. Can you now compute $g'$ using Chain rule?

$\endgroup$

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.