Using the Fund. Theorem of Calc find $h(x)=\int_{2}^\frac{1}{x}\sin^4tdt$'s derivative.

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.

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

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 $$h'(x) = - \frac{1}{x^2} \sin^4 \left( \frac{1}{x} \right).$$
• @AmorFati Also how does $F(2)$ become $\frac{1}{x}$? Dec 10, 2018 at 0:51
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 $$v'(x)=(x^{-1})'=-1x^{-1-1}=-1/x^2$$.