# Need help with a C4 substitution integration question

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#1
I can't seem to get to the right answer. A worked solution or help on what to do would be appreciated, thank you.
0
3 years ago
#2
(Original post by 17fali)
I can't seem to get to the right answer. A worked solution or help on what to do would be appreciated, thank you.
Start by differentiating tan(u) with respect to u - d/du (tanu) = sec^2(u)

Then use Sin^2u + Cos^2u = 1, divide by Cos^2u to find that 1 + tan^2u = Sec^2u

Convert the limits in to limits of u (rather than x). So 3^0.5 becomes tan^-1(3^0.5) and o is still 0.

Then rewrite the integral as Integral between tan^-1(3^0.5) and 0 and (1/secu)*sec^2u . du which is the same as the integral of sec(u) du.

(Multiplying by sec^2(u) so that the integral is with respect to u rather than x)

Either by playing around with logs for a bit or looking in the formula booklet we can see that the integral of sec(u) = ln[sec(u)+tan(u)] (differentiate this to prove it to yourself)

Then substitute the limits in and draw a triangle to find sec(tan^-1(3^0.5)) and simplify to get the answer you are looking for
0
3 years ago
#3
Here's what I think (see attached pic). Don't think there any ln involved. Where did you get the question if you don't mind me asking?
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#4
(Original post by Peter_Don)
Here's what I think (see attached pic). Don't think there any ln involved. Where did you get the question if you don't mind me asking?
Hi, thanks for trying to help. The ln is supposed to be there. I managed to figure it out so I've attached my worked solution in case you wanted to see it The question is from the "Revise Edexcel AS and A Level Modular Mathematics Core Mathematics 4" revision guide from the Edexcel/Pearson website.
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#5
(Original post by RBP_98)
Start by differentiating tan(u) with respect to u - d/du (tanu) = sec^2(u)

Then use Sin^2u + Cos^2u = 1, divide by Cos^2u to find that 1 + tan^2u = Sec^2u

Convert the limits in to limits of u (rather than x). So 3^0.5 becomes tan^-1(3^0.5) and o is still 0.

Then rewrite the integral as Integral between tan^-1(3^0.5) and 0 and (1/secu)*sec^2u . du which is the same as the integral of sec(u) du.

(Multiplying by sec^2(u) so that the integral is with respect to u rather than x)

Either by playing around with logs for a bit or looking in the formula booklet we can see that the integral of sec(u) = ln[sec(u)+tan(u)] (differentiate this to prove it to yourself)

Then substitute the limits in and draw a triangle to find sec(tan^-1(3^0.5)) and simplify to get the answer you are looking for
Thanks so much for helping I managed to do it!
0
3 years ago
#6
(Original post by 17fali)
Thanks so much for helping I managed to do it!
That's good, however when integrating your units should be in radians rather than degrees. It is fine here because the integral only gives trigonometric functions, but if there was a constant term and you had substituted 60* instead of (pi/3) radians you would have got the wrong answer.

Anything else, let me know
0
3 years ago
#7
(Original post by Peter_Don)
Here's what I think (see attached pic). Don't think there any ln involved. Where did you get the question if you don't mind me asking?
The integral of sec(u).du is ln[sec(u) + tan(u)], not sec(u)tan(u). This can be shown by differentiating ln{sec(u)+tan(u)] and then simplifying. I think what you have done is differentiate sec(u).du as d/du (sec(u)) = sec(u)tan(u)
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#8
(Original post by RBP_98)
That's good, however when integrating your units should be in radians rather than degrees. It is fine here because the integral only gives trigonometric functions, but if there was a constant term and you had substituted 60* instead of (pi/3) radians you would have got the wrong answer.

Anything else, let me know
Ah I see, thanks for letting me know that. I will do, thank you!
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