# What is #f(x) = int cosxsinx-sin^2x dx# if #f(pi/6)=1 #?

# f(x) = -1/4cos2x+1/4sin2x -1/2x + 9/8-sqrt(3)/8 +pi/12#

Weh have:

We can use the identities:

Which if applied to the integrand gives us:

Leading to the solution:

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To find ( f(x) = \int \cos(x) \sin(x) - \sin^2(x) , dx ) and given that ( f(\pi/6) = 1 ), you'll need to integrate the function and then evaluate it at ( x = \pi/6 ) to solve for the constant of integration.

[ \int \cos(x) \sin(x) - \sin^2(x) , dx = \frac{1}{2} \sin^2(x) - \frac{1}{3} \sin^3(x) + C ]

Now, we can substitute ( \pi/6 ) into this expression and solve for the constant of integration.

[ 1 = \frac{1}{2} \sin^2(\pi/6) - \frac{1}{3} \sin^3(\pi/6) + C ]

[ 1 = \frac{1}{2} \left( \frac{1}{2} \right)^2 - \frac{1}{3} \left( \frac{1}{2} \right)^3 + C ]

[ 1 = \frac{1}{8} - \frac{1}{24} + C ]

[ 1 = \frac{3}{24} - \frac{1}{24} + C ]

[ 1 = \frac{2}{24} + C ]

[ 1 = \frac{1}{12} + C ]

[ C = 1 - \frac{1}{12} = \frac{11}{12} ]

So, the function ( f(x) = \int \cos(x) \sin(x) - \sin^2(x) , dx ) with ( f(\pi/6) = 1 ) is:

[ f(x) = \frac{1}{2} \sin^2(x) - \frac{1}{3} \sin^3(x) + \frac{11}{12} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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