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I am stuck on this problem:

Find the derivative of the function: $$H(z)= \int_{e^z}^{\cos z} \ln(w^3) ~dw.$$

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Leibniz Integral rule (en.wikipedia.org/wiki/Leibniz_integral_rule) –  user17762 Dec 1 '11 at 4:09
    
Would it help you if we rewrote your function as $$\int_1^{\cos z} \ln(w^3)\mathrm dw-\int_1^{e^z} \ln(w^3) ~dw$$ ? –  J. M. Dec 1 '11 at 4:16

4 Answers 4

up vote 4 down vote accepted

We will use a (very) modified brute force approach. Let $f(w)=\ln(w^3)$. The idea is to find an antiderivative $F(w)$ of $f(w)$. Then $H(z)=F(\cos z)-F(e^z)$. Once we have found $F(w)$, we will have an expression for $H(z)$, and differentiation should be easy.

But we don't need to find $F(w)$! We have $$H(z)=F(\cos z)-F(e^z).$$ Differentiate, using the Chain Rule, and recalling that though we never calculated $F(w)$, we know that $F'(w)=f(w)$. We get $$H'(z)=(-\sin z)f(\cos z)-(e^z)f(e^z).$$ But $f(w)=\ln(w^3)=3\ln w$. Thus $$H'(z)=(-\sin z)(3)(\ln(\cos z)) -(e^z)(3)(z).$$ The answer looks nicer if we write it as $-3\left[\sin z\,\ln(\cos z)+ze^z\right]$.

Comment: In this case, finding $F(w)$, an antiderivative of $\ln(w^3)$, is easy, particularly if we notice that $\ln(w^3)=3\ln w$. However, the technique we used works for "any" $f(w)$, even if we don't know an antiderivative of $f(w)$.

I went through the full reasoning, but one use the reasoning to obtain a general formula, and after that just use the formula. Let $f(w)$, $a(z)$, and $b(z)$ be nice functions, and let $$H(z)=\int_{a(z)}^{b(z)}f(w)\,dw.$$ Then $$H'(z)=b'(z)f(b(z))-a'(z)f(a(z)).$$

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Hint:
If $g(x)$ and $h(x)$ are differentiable and $f(u)$ is a continuous function between $g(x)$ and $h(x)$, then $$\frac{d}{dx}\int_{g(x)}^{h(x)} f(u)~du = f(g(x))\cdot g'(x)-f(h(x))\cdot h'(x)$$

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By the Second Fundamental Theorem of Calculus, if $f(u)$ is continuous we have: $$\frac{d}{dx} \int_a^x f(u)\,du = f(x).$$ Using the Chain Rule, if $h(x)$ is any function, we have: $$\frac{d}{dx} \int_a^{h(x)} f(u)\,du = f(h(x))h'(x).$$

Using this and properties of the integral, we have: $$\begin{align*} \frac{d}{dz}H(z) &= \frac{d}{dz}\int_{e^z}^{\cos z} \ln(w^3)\,dw\\ &=\frac{d}{dz}\left(\int_{e^z}^1\ln(w^3)\,dw +\int_1^{\cos z}\ln(w^3)\,dw\right)\\ &=\frac{d}{dz}\left(\int_1^{\cos z}\ln(w^3)\,dw - \int_1^{e^z}\ln(w^3)\,dw\right)\\ &=\frac{d}{dz}\int_1^{\cos z}\ln(w^3)\,dw - \frac{d}{dz}\int_1^{e^z}\ln(w^3)\,dw\\ &= \ln(\cos^3(z))(\cos z)' - \ln((e^z)^3)(e^z)'\\ &= -3\sin z\ln(\cos z) - 3ze^z. \end{align*}$$

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A brute-force approach:

  1. Use change of variable: $x = w^3$

  2. Use integration by parts on the resulting integral to integrate the function.

  3. Find the derivative the usual way using chain rule/product rules etc.

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2  
For evaluating the integral, it might be slightly easier to observe that $\ln (w^3) = 3 \ln w$, although this approach still does not avoid integration by parts. –  Srivatsan Dec 1 '11 at 4:27

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