Integration!

• Wireless-

  3 Jun 10, 23:15

  7. Find  sqrt(1 + e^2x) dx by the use of a suitable substitution. (12 marks)

   

  PS : I haf the fomulae sheet

• eagle

  3 Jun 10, 23:38

  What level is this maths? You did not state.

  I shall attempt using a a simple method:

  Sub x = ln (tan y)

  Integrate sqrt(1 + e^2x) dx

  = integrate sqrt (1 + e^{2ln (tan y)})   ((1/tan y)( sec²y)) dy

  = integrate sqrt (1 + e^{ln (tan 2y)})   ((1/tan y)( 1 + tan²y)) dy

  = integrate sqrt (1 + tan ²y)   (cot y + tan y) dy

  = integrate sqrt (sec ²y)   (cot y + tan y) dy

  = integrate sec y (cot y + tan y) dy

  = integrate cosec y + tan y sec y dy

  = -ln | cosec y + cot y | + sec y + c

   

  Since tan y = e^x

  cos y = 1 / sqrt(1 + e^2x)

  sin y = e^x / sqrt(1 + e^2x)

   

  -ln | cosec y + cot y | + sec y + c

  = -ln | (sqrt(1 + e^2x) + 1) / e^x | + sqrt(1 + e^2x) + c

  = -ln | (sqrt(1 + e^2x) + 1) | + x + sqrt(1 + e^2x) + c (ans)
  

• Wireless-

  4 Jun 10, 08:12

  Ahhh. eagle thats too chim. Its poly year 2 engineering maths common test

   

   Integrate sqrt(1 + e^2x) dx

• wee_ws

  4 Jun 10, 08:26

  Hi,

    The question is worth 12 marks so the working requires a considerable amount of mathematical effort from the student.

    It is important for you to appreciate why the substitution was chosen for this question, which I will now clarify.

    We are given the expression sqrt [1 + e^(2x) ]. Whenever we see a square root operation, we would like the expression inside to be a perfect square to neutralise the square root operation. We recall in trigonometry that 1 + (tan y)^2 = (sec y)^2. Hence, we take the substitution to be tan y = e^x, which is equivalent to x = ln (tan y) in Eagle's working.

    This is another version:

    Let tan y = e^x.

    Differentiating with respect to x, (sec y)^2 . (dy/dx) = e^x

    => dx/dy = (sec y)^2 / e^x = 1 / (cos y sin y),

    since (sec y)^2 = 1 + (tan y)^2 and e^x = tan y.

    Now int sqrt [ 1 + e^(2x) ] dx = int sqrt [ 1 + (tan y)^2 ] (dx/dy) dy, where we use the chain rule (dx/dy) dy to change the subject of integration from x to y.

    = int (sec y) . {1 / (cos y sin y)} dy, noting that 1 / (cos y sin y) = sec y cosec y.

    = int (sec y)^2 . cosec y dy

    = int {1 + (tan y)^2} . cosec y dy

    = int {cosec y + (tan y)(sec y)} dy

    = -ln |cosec y + cot y| + int { (sin y)(cos y)^(-2) } dy

    = -ln |cosec y + cot y| + sec y + c, which has to be converted back to an expression involving x. We will need to find the other trigonometric ratios with the aid of a right-angled triangle.

    Since tan y = e^x, we have

    opposite side = e^x,

    adjacent side = 1, and

    hypothenuse side = sqrt [ 1 + e^(2x) ] in a right-angled triangle, which you should draw to visualise.

    From here, we can find expressions for cosec y (ratio is hypothenuse/opposite), cot y (ratio is adjacent/opposite) and sec y (ratio is hypothenuse/adjacent), so that our final answer can be represented in terms of x, as Eagle has provided.

    Thanks.

  Cheers,
  Wen Shih

• eagle

  4 Jun 10, 08:30

  oops... I made it a bit too unclear... my bad... :(

  Yup Mr Wee is right.

  tan y = e^x would be easier to see.

• Wireless-

  4 Jun 10, 23:12

  wow pro. it makes alot sense now lol.Teacher nvr teach this type of substitutions b4. thanks for teaching dudes!

   

  I have a qn.

   

    = int [(sec y)^2 . cosec y ]dy             <=This part

  what if instead of

  = int (sec y)^2 . cosec y dy

    = int {1 + (tan y)^2} . cosec y dy

    = int {cosec y + (tan y)(sec y)} dy

   

  doing this way.

   

  I did this way instead and got stuck.

   

    = int [(sec y)^2 . cosec y ]dy

  =int[(1/cos y)^2. (1/sin y)]dy

  =int[(1/[(cosy)^2  *(sin y)] dy

   

  Stucked here. Can that be further integrated? or have to do it ur way?

• wee_ws

  5 Jun 10, 06:21

  Hi,

    This type of substitution requires critical thinking as well as appreciating the approach to deal with expressions involving the square root operator. I believe your lecturer would have covered (at least) the substitutions of x = sin u or x = cos u when integrating the expression sqrt(1 - x^2) and the substitution x = tan u when tackling the integral of sqrt(1 + x^2).

    It can't be integrated further unless we recognise 1/(cos y)^2 as (sec y)^2, which can then be rewritten as 1 + (tan y)^2, from which we proceed as discussed previously.

    Whenever trigonometry is involved, one needs to be well-versed in the various trigonometric results and to be highly flexible in manipulations to make the expression easy to integrate.

    Thanks.

  Cheers,
  Wen Shih

• Mad Hat

  5 Jun 10, 11:59

  Re: Wireless. Where you got stuck, you could also have integrated by parts.
  Choose u = cosec y , v ' = sec² y.

  ∫ sec² y cosec y dy
  =  cosec y  tan y  + ∫ cosec y dy
  =  sec y + ∫ cosec y dy

  which is the same result as eagle and Mr. Wee.

  But the question is tough. Finding the correct substitution would be challenging to most students and also ∫ cosec , needed at the end, is not exactly the most famous integral.

  No wonder 12 marks.