How do you find the antiderivative of $\int ({cos}^{2} x - {sin}^{2} x) d x$ $int (cos^2x-sin^2x)dx$ from $[0, \frac{π}{6}]$ $[0,pi/6]$ ?

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Answer 1 · 2016-10-19T19:43:44

$\int_{0}^{\frac{π}{6}} ({cos}^{2} x - {sin}^{2} x) d x = \frac{\sqrt{3}}{4}$ $int_0^(pi/6)(cos^2x-sin^2x)dx=sqrt3/4$

$\int_{0}^{\frac{π}{6}} ({cos}^{2} x - {sin}^{2} x) d x$ $int_0^(pi/6)(cos^2x-sin^2x)dx$

$= \int_{0}^{\frac{π}{6}} (cos 2 x) d x$ $=int_0^(pi/6)(cos2x)dx$

$= {[\frac{1}{2} sin 2 x]}_{0}^{\frac{π}{6}}$ $=[1/2sin2x]_0^(pi/6)$

$= \frac{1}{2} sin 2 (\frac{π}{6}) - \frac{1}{2} sin 2 (0)$ $=1/2sin2(pi/6)-1/2sin2(0)$

$= \frac{\sqrt{3}}{4} - 0 = \frac{\sqrt{3}}{4}$ $=sqrt3/4-0=sqrt3/4$

How do you find the antiderivative of int (cos^2x-sin^2x)dx∫(cos2x−sin2x)dxint (cos^2x-sin^2x)dx from [0,pi/6][0,π6][0,pi/6]?