Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$.

Question

$$\begin{align} \left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\ &= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\ &= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\ \end{align}$$

Update:

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$$\begin{align} \left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\ &= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\ \end{align}$$

Is this correct?

It seems like you don't have any clue what integration is... — , Mar 30 '15 at 16:10
Your answer is indeed far from correct..... Can you explain to us how you get, for example, the first equality? — , Mar 30 '15 at 16:10
You probably mean differentiate. Then you must replace $t$ by $x^2$ in the first part and by $\tan x$ in the second part. — André Nicolas, Mar 30 '15 at 16:10
No, and I suggest you to try to compute something simpler, like $\int_0^x t,dt$ (the same principle), if you don't know how to do... — mickep, Mar 30 '15 at 16:10
I originally had "integrate" when that was incorrect, I edited to to read "Find the derivative". — KKendall, Mar 30 '15 at 16:18
@AndréNicolas thank you for the help. That is what I meant. I was confusing the two. — KKendall, Mar 30 '15 at 16:37

randomgirl · Answer 1 · 2015-03-30T16:38:40.940

3

$\text{ Recall } \int_{a(x)}^{b(x)}f(t) dt=F(b(x))-F(a(x)) \text{ *note:} \text{( where } F'=f) \\ \frac{d}{dx}[F(b(x))-F(a(x))]=\frac{d}{dx}F(b(x))-\frac{d}{dx}F(a(x)) \text{ by difference rule } \\ \\ \text{ now you use chain rule and you are almost there } $

edited Mar 30 '15 at 16:38

answered Mar 30 '15 at 16:25

randomgirl

3,237

1

You should use dummy variables, for example $\int_{a(x)}^{b(x)} f(y) dy$. The integration variable doesn't "exist" outside of the integration scope. – Ian Mar 30 '15 at 16:37
Yeah I usually do but I'm typing around a cat :p and thanks – randomgirl Mar 30 '15 at 16:38

Trajan · Answer 2 · 2015-03-30T16:23:17.000

Hint to find derivative: Use FTC, http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements, $F'(x)=\frac{d}{dx}\int^x_a f(t)dt=f(x)$

Answer to original question: HINT to integrate: \begin{align} \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt &= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\ &= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\ \end{align} Then let $\tan^2 u=t^4/2$

score 0 · Answer 3 · answered Mar 30 '15 at 16:10

0

No. $t$ is a mute variable. The answer must have $x$ as the only variable. Yor computations are similar to $$ \int_1^2t\,dt=2\cdot t-1\cdot t=t, $$ which is obviously wrong.

answered Mar 30 '15 at 16:10

Julián Aguirre

76,354

Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$.

3 Answers3