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Consider the integral $$h(x)=\int_{2}^\frac{1}{x}\sin^4(t)dt$$

In my notes I have the integral equal to $$\sin^4\left(-\frac{1}{x^2}\right)\cdot\frac{1}{x}$$

and the following answer as $$h'(x)=-\sin^4\frac{1}{x}/{x^2}$$

Obviously I skipped steps and looked up the answer in the back of the book, would really appreciate it if someone could fill me in.

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  • $\begingroup$ You should explain also what you have tried so far. $\endgroup$
    – user
    Dec 10, 2018 at 0:38

2 Answers 2

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Set $\mathscr{F}'(x) : = \sin^4(x)$, i.e., $\mathscr{F}$ is the antiderivative of $\sin^4(x)$. Then by the fundamental theorem of calculus $$h(x) = \mathscr{F} \left( \frac{1}{x} \right) - \mathscr{F}\left( 2 \right).$$ Differentiating this expression, we see that $$h'(x) = -\frac{1}{x^2} \mathscr{F}' \left( \frac{1}{x} \right).$$ Since $\mathscr{F}'(x) = \sin^4(x)$, it follows that $$h'(x) = - \frac{1}{x^2} \sin^4 \left( \frac{1}{x} \right).$$

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  • $\begingroup$ As a suggetion, it should be better for the asker and for the site do not give full answer when question are posed in that way. I think a good hint or a guide would be much more useful. $\endgroup$
    – user
    Dec 10, 2018 at 0:45
  • $\begingroup$ I agree, but having not studied this topic in a while I had no idea on how to advance. $\endgroup$
    – Eric Brown
    Dec 10, 2018 at 0:48
  • $\begingroup$ @AmorFati Also how does $F(2)$ become $\frac{1}{x}$? $\endgroup$
    – Eric Brown
    Dec 10, 2018 at 0:51
  • $\begingroup$ @gimusi I feel this is sufficiently elementary to offer a complete solution. $\endgroup$
    – AmorFati
    Dec 10, 2018 at 0:51
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    $\begingroup$ @AmorFati Ahhh I see now, thanks for your help. $\endgroup$
    – Eric Brown
    Dec 10, 2018 at 0:52
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Say we had a function $F$ so that $F'(t)=\sin^4t$. Then $h(x)=F(1/x)-F(2)$, so $$h'(x) =^* F'\left(\frac1x\right)\left(-\frac1{x^2}\right)=-\frac1{x^2}\left(\sin^4\left(\frac1x\right)\right).$$ Note that the expression after $=^*$ is derived by applying both the chain rule and the power rule. The chain rule comes in as $\frac d{dx}u(v(x))=u'(v(x))v'(x)$ where $u(v)=F(v)$, and $v(x)=1/x$. The power rule comes in when calculating $v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$.

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