Have you learnt about trigonometric substitutions - yet? (n/t)


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Posted by Subhotosh Khan on August 15, 2002 at 08:46:27:

In Reply to: help! posted by Andrew Cash on August 14, 2002 at 19:04:26:

: I am having extreme difficulty integrating rational expressions with x terms in the denominator. Can anyone help me with both an answer AND an explanation.
: INTEGRATE: 1 / x((x^2 - 4)^1/2)
: and
: INTEGRATE: 1 / (x^1/2)(1 - 2(x^1/2))
: Thankyou for taking the time.

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These problems - GENERALLY - become easier to deal with after trigonometric substitution:

INT[1 / x((x^2 - 4)^1/2)] dx

substitute:

x = 2*sec T ...................(1)
dx = 2*sec T * tan T dT

INT[1 / x((x^2 - 4)^1/2)] dx

= INT[1/(2*sec T) * ((4 sec^2 T - 4)^1/2)]*2*sec T * tan T dT

= INT[1/(2*sec T) * ( 2*tan T)]*2*sec T * tan T dT

= INT[ 2*tan^2 T] dT

=2*INT[ sec^2 T - 1] dT

=2*[ tan T - T] + C

Now replace 'T' by 'arccos(1/2x)'{...from (1)}

The second one - does not need trigonometric sustitution. You canreduce it algebraically ---

1 / (x^1/2) (1 - 2(x^1/2))

= x^(-1/2) - 2

Now this should be easy....


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