How do you find the derivative of $y= int 6*(sin(t))^2 dt$ from a=e^x to b=1?

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Answer 1 · 2016-01-10T09:27:28

Use second fundamental theorem of calculus to find the answer. The working is given below.

$y=int_(e^x)^1 6*(sin^2(t))dt$
$y=-int_1^(e^x) 6*(sin^2(t))dt$

$dy/dx = -dy/dx(int_1^(e^x) 6*(sin^2(t))dt)$

Using second fundamental theorem of calculus we replace $t$ by $e^x$ and multiply the result by derivative of $e^x$

$dy/dx = -(6sin^2(e^x))(e^x)$

$dy/dx = -6e^xsin^2(e^x)$ Answer

How do you find the derivative of y= int 6*(sin(t))^2 dty= int 6*(sin(t))^2 dt from a=e^x to b=1?