Prove that -cos 2 x dx - x 2 -sin 2x 4 -c-

∫cos ^ 2 x #160; dx #160; =∫ 1+cos2x / 2 dx #160; =∫( 1 / 2 #160;+cos2x / 2 ) dx #160; = x 2 +sin2x 4 +c , #160;where ...

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Second Fundamental Theorem of Calculus

Prove that ...

Question

Prove that $\int {cos}^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4} + c$ .

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\int \frac{1 - sin 2 x}{x + {cos}^{2} x} d x

{cos}^{2} (x) + {cos}^{2} (x + \frac{π}{3}) + {cos}^{2} (x - \frac{π}{3}) = \frac{3}{2}

\int \frac{{sin}^{3} x d x}{(1 + {cos}^{2} x) \sqrt{1 + {cos}^{2} x + {cos}^{4} x}} d x

I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{sec x}}{\sqrt{cosec x} + \sqrt{sec x}} d x = \frac{π}{4}

\frac{1 + s i n 2 x}{c o s 2 x} = t a n (\frac{π}{4} + x)

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