# How do you use the second fundamental theorem of Calculus to find the derivative of given #int ((sin^3)(t))dt# from #[0, e^x]#?

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To use the Second Fundamental Theorem of Calculus to find the derivative of ( \int_{0}^{e^x} \sin^3(t) , dt ), first, you need to find the antiderivative of the integrand, ( \sin^3(t) ). Then, apply the Second Fundamental Theorem of Calculus, which states that if ( f ) is continuous on ([a, b]) and ( F ) is an antiderivative of ( f ) on ([a, b]), then ( \int_{a}^{b} f(x) , dx = F(b) - F(a) ).

The antiderivative of ( \sin^3(t) ) can be found by using the power reduction formula for trigonometric functions. The power reduction formula for ( \sin^3(t) ) is ( \sin^3(t) = \frac{3\sin(t) - \sin(3t)}{4} ).

Now, integrate this expression with respect to ( t ) from ( 0 ) to ( e^x ):

[ \int_{0}^{e^x} \frac{3\sin(t) - \sin(3t)}{4} , dt ]

After integrating and evaluating at the limits of integration, you'll obtain the derivative of the given integral with respect to ( x ).

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To find the derivative of the integral ( \int_{0}^{e^x} \sin^3(t) , dt ) with respect to ( x ), we can apply the second fundamental theorem of calculus. According to this theorem, if ( F(x) ) is a continuous function on the interval ([a, b]), and ( f(x) ) is a continuous function on an interval containing ([a, b]), then:

[ \frac{d}{dx} \left( \int_{a}^{u(x)} f(t) , dt \right) = f(u(x)) \cdot \frac{du}{dx} ]

Given that ( u(x) = e^x ) and ( f(t) = \sin^3(t) ), we can proceed as follows:

[ \frac{d}{dx} \left( \int_{0}^{e^x} \sin^3(t) , dt \right) = \sin^3(e^x) \cdot \frac{d}{dx}(e^x) ]

[ = \sin^3(e^x) \cdot e^x ]

Therefore, the derivative of the given integral ( \int_{0}^{e^x} \sin^3(t) , dt ) with respect to ( x ) is ( \sin^3(e^x) \cdot e^x ).

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