find the integral of y=x^2 +sin^2(x) with respect to x between the limits 0 and pi

intergal0pi(x^2 + sin^2(x))dx = integral0pi(x^2 + 1/2 - cos(2x)/2)dx since sin^2(x) + cos^2(x) = 1cos^2(x) - sin^2(x) = cos(2x)2sin^2(x) = 1 - cos(2x)thereforeintegral0pi(x^2 +sin^2(x))dx = [(x^3)/3 + x/2 - sin(2x)/4]0pi= ((pi^3)/3 + pi/2 - 0/4) - (0)=(pi^3)/3 + pi/2

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The curve C has equation y=2x^2 -11x +13. (a) The point P has coordinates (2, – 1) and lies on C. Find the equation of the tangent to C at P.


Express 6cos(2x)+sin(x) in terms of sin(x). Hence solve the equation 6cos(2x) + sin(x) = 0, for 0° <= x <= 360°.


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