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Find the derivative of the function $$y=\int_{\cos x}^{\sin x}\ln(3+7v)\mathrm dv.$$

I know it is supposed to use the FTC in some way.

When I got $\cos(x) \ln(3) + 5\sin(x) + \sin(x) \ln(4) + 5\cos(x)$ the answer was incorrect.

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    $\begingroup$ Could you add the work you did prior to arriving at what you "got"? $\endgroup$ – Namaste Jul 15 '18 at 17:19
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Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is

$$I(x)=F(\sin x)-F(\cos x).$$

Now by the chain rule,

$$I'(x)=(F(\sin x))'-(F(\cos x))'=f(\sin x)(\sin x)'-f(\cos x)(\cos x)' \\=\ln(3+7\sin x)\cos x+\ln(3+7\cos x)\sin x.$$

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  • $\begingroup$ +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer. $\endgroup$ – Cameron Buie Jul 15 '18 at 19:38
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There is a more general formula, assuming all functions are $C^1$:

$$\frac{d}{dx} \int_{b(x)}^{a(x)} f(x,v)\,dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + \int_{g(x)}^{h(x)} \frac{\partial }{\partial x}f(x,v)\,dv$$

We immediately get

$$\frac{d}{dx} \int_{\cos x}^{\sin x}\ln(3+7v)\,dv = \ln(3+7\sin(x))\cos(x)+\ln(3+7\cos(x))\sin(x)$$

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