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

Question

Charles Arocha · Accepted Answer

# 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:

#x^2 = 4 sinh^2 y#

#implies 2 x \ dx = 8 sinh y \ cosh y \ dy#

#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 #

Ezra Adams · Answer

undefined To evaluate the integral ∫√(4 + x^2), you can use trigonometric substitution. Let x = 2tan(θ), then dx = 2sec^2(θ)dθ. Substituting these into the integral gives ∫√(4 + (2tan(θ))^2) * 2sec^2(θ)dθ. Simplify this to get ∫√(4 + 4tan^2(θ)) * 2sec^2(θ)dθ. Using the trigonometric identity sec^2(θ) = 1 + tan^2(θ), you get ∫√(4sec^2(θ)) * 2sec^2(θ)dθ. This simplifies to ∫2(2sec(θ)) * 2sec^2(θ)dθ. Now, simplify further to get ∫4sec^3(θ)dθ. This integral can be evaluated using the formula for the integral of sec^3(θ), which is (1/2)sec(θ)tan(θ) + (1/2)ln|sec(θ) + tan(θ)| + C. Substitute back θ = arctan(x/2) to get the final answer.