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# fp2 - integrating inverse trig watch

1. Hi,

I need to evaluate:

My solution:

Let x = 5u so

changing the limits: when x=0, u=0 and when x=2,u=10

(Sorry, I can't quite get the last 2 integrals to look right, the weird bit at the beginning of each integral is supposed to say the integral between 10 and 0 of (fraction) du )

= [(1/5) tan^-1 (u)] between 0 and 10

= (1/5) tan^-1 (10) - (1/5) tan^-1 (0)

= 0.294 (3sf)

but the answer should be 0.0761. I think the problem is to do with the new limits, ie. the 10 should be 10/25 = 2/5 but I'm not sure why. Any help would be much appreciated. Thanks
2. if x = 5u then when x = 2 surely u = 2/5?
3. (Original post by Laurenje)
Hi,

I need to evaluate:

My solution:

Let x = 5u so

changing the limits: when x=0, u=0 and when x=2,u=10

(Sorry, I can't quite get the last 2 integrals to look right, the weird bit at the beginning of each integral is supposed to say the integral between 10 and 0 of (fraction) du )

= [(1/5) tan^-1 (u)] between 0 and 10

= (1/5) tan^-1 (10) - (1/5) tan^-1 (0)

= 0.294 (3sf)

but the answer should be 0.0761. I think the problem is to do with the new limits, ie. the 10 should be 10/25 = 2/5 but I'm not sure why. Any help would be much appreciated. Thanks
x=5u so u=0.2x. The new limits are one fifth of the originals, not five times.
4. Oops!! That's so typical of me, I do the hard bit, but then fail at the basics. Thank you.

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