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вершины y=(-3x^2-2x+6)/5

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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∫ tan(5x)dx

Solution

−15 ln|cos(5x)|+C

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∫ tan(5x)dx

Перепишите используя тригонометрические тождества

=∫ sin(5x)cos(5x) dx

Применить подстановку интеграла: ∫ −15u du

=∫ −15u du

Извлечь константу: ∫a·f(x)dx=a·∫f(x)dx

=−15 · ∫ 1u du

Использовать общий интеграл: ∫ 1u du=ln(|u|)

=−15 ln|u|

Делаем обратную замену u=cos(5x)

=−15 ln|cos(5x)|

Добавить константу к решению

=−15 ln|cos(5x)|+C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

вершины y=(-3x^2-2x+6)/5

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