How do you evaluate the integral #int sqrt(4+x^2)#?

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Answer 1 · 2018-05-29T22:04:57

# I = x sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C #

#I = int sqrt(4+x^2) color(red)(dx) #

You can use identity:

So let:

#implies I = int sqrt(4+4 sinh^2 y) \ (8 sinh y \ cosh y \ dy)/(2x)#

# = int 2 cosh y \ (8 sinh y \ cosh y \ dy)/(2* 2 sinh y)#

# =4 int cosh^2 y \ dy #

# =4 int (cosh 2y +1 ) /2\ dy #

# = sinh 2y + 2 y + C qquad triangle #

Considering:

#color(red)(sinh 2y = 2 sinh y cosh y )#
#x^2 = 4 sinh^2 y implies color(red)( x = 2 sinh y) color(red)(implies y = sinh^(-1) (x/2)) #
# cosh^2y = 1 + sinh^2 y implies color(red)(cosh y = sqrt(1 + x^2/4) )#

Then #triangle# becomes:

# I = 2 * x/2 * sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C #

# = x sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C #