How to solve the integration question #int1/(4+9x^2)^(1/2)# using trigonometric substitution?

#int1/(4+9x^2)^(1/2)#

Answer 1

#int dx/(4+9x^2)^(1/2) = 1/3 ln abs ( (sqrt(4+9x^2)+3x))+ C#

Substitute:

#x = 2/3tant#
#dx = (2dt)/(3cos^2t) dt#
As then #t = arctan ((3x)/2)# we have that #t in (-pi/2,pi/2)#

we have:

#int dx/(4+9x^2)^(1/2) = 2/3 int ( dt)/cos^2t 1/(4+9*4/9tan^2t)^(1/2) = 1/3 int ( dt)/cos^2t 1/(1+tan^2t)^(1/2)#

Use now the trigonometric identity:

#1+tan^2t = 1+sin^2t/cos^2t = (cos^2t+sin^2t)/cos^2t = 1/cos^2t#
#int dx/(4+9x^2)^(1/2) = 2/3int ( dt)/cos^2t 1/(1/cos^2t)^(1/2)#
For #t in (-pi/2,pi/2)#, #cos t >0# so:
#(1/cos^2t)^(1/2) = 1/cost#

and:

#int dx/(4+9x^2)^(1/2) = 2/3int ( dt)/cos^2t 1/(1/cost) = 2/3int (costdt)/cos^2t#

we do not simplify but substitute again:

#u=sint#
#du = cost dt#
#cos^2t = 1-sin^2t = 1-u^2#
#int dx/(4+9x^2)^(1/2) = 2/3 int (du)/(1-u^2)#

which can be solved by partial fractions:

#1/(1-u^2) = 1/((1+u)(1-u)) = 1/2 1/(1+u) + 1/2 1/(1-u)#

so:

#int (du)/(1-u^2) = 1/2int (du)/(1-u) +1/2 int (du)/(1+u) = -1/2lnabs(1-u)+1/2 ln abs (1+u)+C#

and using the properties of logarithms:

#int (du)/(1-u^2) = ln abs ((1+u)/(1-u))^(1/2) + C#

Reversing the substitution we have:

#u = sint = tant cost = tant/sqrt(1+tan^2t) = (3/2x)/sqrt(1+(3/2x)^2) = (3x)/sqrt(4+9x^2)#

So:

#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (1 +((3x)/sqrt(4+9x^2)))/ ( 1 -((3x)/sqrt(4+9x^2)))) + C#
#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)/( sqrt(4+9x^2) - 3x))+ C#

Rationalize the denominator of the argument:

#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( ((sqrt(4+9x^2)+3x)( sqrt(4+9x^2) + 3x))/ ( ( sqrt(4+9x^2) - 3x)( sqrt(4+9x^2) + 3x)))+ C#
#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( ((sqrt(4+9x^2)+3x)( sqrt(4+9x^2) + 3x))/ ( (4+9x^2-9x^2))+ C#
#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)^2)/ sqrt( (4+9x^2-9x^2))+ C#
#int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)^2)/ sqrt( 4)+ C#
#int dx/(4+9x^2)^(1/2) = 1/3 ln abs ( (sqrt(4+9x^2)+3x))+ C#
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Answer 2

To solve the integral ∫(1 / √(4 + 9x^2)) dx using trigonometric substitution, follow these steps:

  1. Recognize the form of the integral and choose the appropriate trigonometric substitution. In this case, since we have 4 + 9x^2, it suggests using x = (2/3)tan(θ).

  2. Compute dx in terms of dθ using the derivative of the substitution x = (2/3)tan(θ).

  3. Substitute x and dx in terms of θ into the integral.

  4. Simplify the integrand using trigonometric identities to express it solely in terms of θ.

  5. Integrate the expression with respect to θ.

  6. Finally, revert back to the original variable x using the trigonometric identities and simplify the result.

By following these steps, you can solve the given integral using trigonometric substitution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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