How do you integrate #int 1/sqrt(e^(2x)-2e^x+10)dx# using trigonometric substitution?

Answer 1
#int 1/sqrt(e^(2x)-2e^x+10)dx#
#int (e^xe^-x)/sqrt(e^(2x)-2e^x+10)dx#
#t = e^x# #dt = e^x#
#int 1/(tsqrt(t^2-2t+10))dt#
#int 1/(tsqrt(t^2-2t+1 + 9))dt#
#int 1/(tsqrt((t-1)^2 + 9))dt#
#t-1 = 3tan(u)# #dt = 3(tan^2(u)+1)du#
#(t-1)^2 = 9tan^2(u)#
#int 1/(tsqrt((t-1)^2 + 9))dx#
#int(3/cos^2(u))/((3tan(u)+1)(3/cos(u))) du#
#int(3/cos^2(u))/((9sin(u)/cos^2(u)+3/cos(u))) du#
#int1/((3sin(u)+cos(u))) du#
#w = tan(u/2)#
#3sin(u) = (6w)/(w^2+1)#
#cos(u) = (1-w^2)/(w^2+1)#
#du = (2dw)/(w^2+1) #
#int(2/(w^2+1))/(((6w)/(w^2+1)+(1-w^2)/(w^2+1))) dw#
#int2/(6w+1-w^2) dw#
#-int2/(w^2-6w+9-10) dw#
#-2int1/((w-3)^2-10) dw#
#sqrt(10)v = (w-3)# #sqrt(10)dv = dw#
#-2sqrt(10)/10int1/(v^2-1) dv#
#-2sqrt(10)/10[arctanh(v)]+C#
#-2sqrt(10)/10[arctanh((tan(arctan((e^x-1)/3)/2)-3)/sqrt(10))]+C#
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Answer 2

To integrate ( \int \frac{1}{\sqrt{e^{2x} - 2e^x + 10}} , dx ) using trigonometric substitution, follow these steps:

  1. Recognize that the expression inside the square root resembles a quadratic equation in terms of ( e^x ).
  2. Let ( e^x = 3 \sec(\theta) ), so ( dx = 3 \sec(\theta) \tan(\theta) , d\theta ).
  3. Substitute ( e^x = 3 \sec(\theta) ) and ( dx = 3 \sec(\theta) \tan(\theta) , d\theta ) into the integral.
  4. Simplify the expression under the square root using the trigonometric identity ( \sec^2(\theta) - 1 = \tan^2(\theta) ).
  5. After simplification, the integral becomes ( \int \frac{1}{3\sqrt{\tan^2(\theta) + 9}} \cdot 3\sec(\theta) \tan(\theta) , d\theta ).
  6. Simplify further to get ( \int \frac{\sec(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta) + 9}} , d\theta ).
  7. Use the trigonometric identity ( \sec(\theta) \tan(\theta) = \frac{d}{d\theta}(\tan(\theta) + 3) ).
  8. Now integrate ( \frac{\tan(\theta) + 3}{\sqrt{\tan^2(\theta) + 9}} , d\theta ) with respect to ( \theta ).
  9. Finally, substitute ( e^x ) back in terms of ( \theta ) and simplify the expression.

This process should lead you to the integral of the original expression 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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