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periodicity of cos(2t)

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∫ √2x²+5dx

Solution

12 x√5+2x²+52√2 ln|√25 x+√15 (5+2x²)|+C

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∫ √2x²+5dx

Apply Trigonometric Substitution: ∫ 5√2 sec³(u)du

=∫ 5√2 sec³(u)du

Take the constant out: ∫a·f(x)dx=a·∫f(x)dx

=5√2 · ∫ sec³(u)du

Apply Integral Reduction

=5√2 (sec²(u)sin(u)2 +12 · ∫ sec(u)du)

∫ sec(u)du=ln|tan(u)+sec(u)|

=5√2 (sec²(u)sin(u)2 +12 ln|tan(u)+sec(u)|)

Substitute back u=arctan(√2√5 x)

=5√2 (sec²(arctan(√2√5 x))sin(arctan(√2√5 x))2 +12 ln|tan(arctan(√2√5 x))+sec(arctan(√2√5 x))|)

Simplify 5√2 (sec²(arctan(√2√5 x))sin(arctan(√2√5 x))2 +12 ln|tan(arctan(√2√5 x))+sec(arctan(√2√5 x))|): 12 x√5+2x²+52√2 ln|√25 x+√15 (5+2x²)|

=12 x√5+2x²+52√2 ln|√25 x+√15 (5+2x²)|

Add a constant to the solution

=12 x√5+2x²+52√2 ln|√25 x+√15 (5+2x²)|+C

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

periodicity of cos(2t)

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