# Need help with a C4 substitution integration question

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I can't seem to get to the right answer. A worked solution or help on what to do would be appreciated, thank you.

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I can't seem to get to the right answer. A worked solution or help on what to do would be appreciated, thank you.

**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.

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

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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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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?

**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?

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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

**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

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Thanks so much for helping I managed to do it!

**17fali**)Thanks so much for helping I managed to do it!

Anything else, let me know

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**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?

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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

**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

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