Question

Найдите первообразную Даю 31 балл!
$y = - 2 { \cos }^{2} (2x)$
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Answer 1

$\int -2cos^2(2x)dx=-2\int cos^2(2x)dx=\begin{vmatrix}2x=t & \\  dx=\frac{dt}{2}& \end{vmatrix}=\\=-\int cos^2(t)dt=-\int \frac{cos(2t)+1}{2}dt=\\=-\frac{1}{2}\int cos(2t)dt-\frac{1}{2}\int dt=-\frac{sin(2t)}{4}-\frac{t}{2}\Rightarrow \\\Rightarrow -\frac{sin(4x)}{4}-x+C$

Answer 2

$\int (-2cos^22x)\, dx=-2\int \frac{1+cos4x}{2}\, dx=-\int (1+cos4x)\, dx=\\\\=-\int dx-\frac{1}{4}\int cos4x\cdot \underbrace {4\, dx}_{d(4x)}=-x-\frac{1}{4}\, sin4x+C$