Question

Решите только примеры 2 и 4.

Answer 1

$2) \: \: 0.2\tg{5x}{ - }cos(x - 2) + C ; \: \: C\in R$

$4)\: \: \sqrt{2x} + 1.5 \tg{2x} + C; \: \: C\in R$

Объяснение:

2)

$\: \: \: \: \int \bigg( \sin(2 - x) + \frac{1}{ \cos^{2} 5x}\bigg){dx} = \\ = \int( \sin(2 - x) dx + \int\frac{1}{ \cos^{2} 5x}dx = \\ \small = \int \big( - \sin(x - 2 )\big) d(x - 2 ) + \int\frac{ \tfrac{1}{5} d(5x)}{ \cos^{2} 5x}= \\ \small = - \int( \sin(x - 2) d(x - 2)+ \frac{1}{5} \int\frac{d(5x)}{ \cos^{2} 5x}dx = \\ = - cos(x - 2) + \frac{1}{5} \tg{5x} + C =\\ = 0.2\tg{5x} - cos(x - 2) + C ; \: \: C\in R$

4)

$\:\: \int\bigg( \frac{1}{ \sqrt{2x} } - \frac{3}{ \sin^{2} 2x } \bigg)dx= \\ \small = \int{\frac{dx}{ \sqrt{2x} }}{ - } \int \frac{3dx}{ \sin^{2} 2x } = \\ \small= \frac{1}{2} \int(2x)^{ - 0.5} {d(2x)} - \frac{3}{2} \int {\frac{d(2x)}{ \sin^{2} 2x }} = \\ \ \frac{1}{2} \cdot \frac{( {2x})^{0.5} }{ 0.5} - \frac{3}{2} \cdot( - \tg{2x}) + C = \\ = \sqrt{2x} + 1.5 \tg{2x} + C; \: \: C\in R$