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I was wondering what the answer to this question is. I just had a test and I want to make sure I was correct. So I need to find the derivative of $ \int_{x^6}^{0} \cos(\sqrt{t}) ~ dt $ (I believe it is $dt$ however I may be wrong and it could have been $dx$.)

The answer I thought was $-3x^2 \cos(x^3)$ but now I realize the answer might be $-6x^5 \cos(x^3)$. Thank you in advance!

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    You should post your answer.2011-12-06

3 Answers 3

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Here's one way to do this kind of problem (without Wolfram).

Let $F(t)=\int\cos(\sqrt t)\,dt$, so F'(t)=\cos(\sqrt t) Then $\int_{x^6}^0\cos(\sqrt t)\,dt=F(0)-F(x^6)$ So the derivative is (F(0))'-(F(x^6))'. Well, $F(0)$ is a constant, so its derivative is zero, so we just have to figure out -(F(x^6))'. By the chain rule, this is -F'(x^6)(x^6)'=-\cos(\sqrt{x^6})(6x^5)=-6x^5\cos|x^3|=-6x^5\cos(x^3) since cosine is an even function.

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    "This answer is useful" - especially for those who aren't allowed to access W|A during tests.2011-12-07
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By the Fundamental Theorem of Calculus and the chain rule,

If $F(x) = \int_{a}^{g(x)}f(t)dt$, then F\ '(x) = f(g(x)) \cdot g'(x).

In particular, for $f(t) = \cos(\sqrt{t}); g(x) = x^6$, we have F \ '(x) = 6x^5 \cdot \cos(x^3)

(with simplifications as in Gerry Myerson's answer).

But then we have that $\int_a^b \text{(stuff)} = - \int_b^a \text{(stuff)}$, which gives us the "$-$" in front as desired.

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As you correctly realized the right answer is $-6x^5\cos(x^3)$.

However, just for confirming such results you don't need to ask us, there are efficient means to help yourself:

Note that the latter can already be achieved by just pressing the result in the first link.