How do you use the Fundamental Theorem of Calculus to find the derivative of #int 5(sin(t))^5 dt# from 5 to e^x?

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Answer 1 · 2016-06-25T20:08:49

#int_5^(e^x)5sin^5tdt=-5{cose^x-2/3cos^3e^x+1/5cos^5e^x}+5{cos5-2/3cos^3(5)+1/5cos^5(5)}.#

Fundamental Theorem of Calculus : #intf(x)dx=F(x)+C rArr int_a^bf(x)dx=[F(x)]_a^b=F(b) -F(a)#

So, let us first find #I=int5sin^5tdt=int5sin^4t*sintdt=int5(1-cos^2t)^2*sintdt............(1).#

Now take substitution #cost=y,# so that #-sintdt=dy.

Hence, #I=5int(1-y^2)^2*(-dy)=-5int(1-2y^2+y^4)dy=-5(y-2/3y^3+1/5y^5)+C =-5(cost-2/3cos^3t+1/5cos^5t)+C.#

Finally, #int_5^(e^x)5sin^5tdt=-5{cose^x-2/3cos^3e^x+1/5cos^5e^x}+5{cos5-2/3cos^3(5)+1/5cos^5(5)}.#

Answer 2 · 2016-07-09T18:18:40

#= 5(sin(e^x))^5 * e^x#

we want #d/dx int_5^{e^x} 5(sin(t))^5 dt# using FTC

FTC tells us that #d/(du) int_a^u f(t) dt = f(u)#

here a = 5, u = e^x

so FTC + chain rules tells us that

#d/(dx) int_a^u f(t) dt = d/(du) int_a^u f(t) dt * (du)/dx = f(u)* (du)/dx#

#= 5(sin(u))^5 * e^x#

#= 5(sin(e^x))^5 * e^x#