# integration by substitutionWatch

This discussion is closed.
#1
Use the substitution x=sin theta to show that:

1 / (1-x^2)^3/2 dx = x/(1-x^2)^1/2 + c

Plz can someone help me!
0
13 years ago
#2
dx/dt = cost
dx = cost dt

=> I = INT [1/(1-sin2t)3/2].[cost]dt

= INT cost/cos3t dt

= INT sec2t dt

= tant + c

Draw a right-angled triangle, with angle t.
The opposite side is x, and the hypotenuse is 1 (since sint = x). The remaining side is (1-x2)1/2
=> tant = x/(1-x2)1/2

=> I = x/(1-x2)1/2 + c
0
13 years ago
#3
int 1 / (1-x^2)^3/2 dx
x = [email protected]
dx/[email protected] = [email protected]

int 1 / (1-x^2)^3/2 dx = int 1 / (1-sin²@)^3/2 dx
= int 1 / (cos²@)^3/2 dx
= int [1 / (cos²@)^3/2 ][email protected] [email protected]
= int [1 / (cos³@) ][email protected] [email protected]
= int 1/cos²@ [email protected]
= int sec²@ [email protected]
= [email protected] + C

now x = [email protected]
so tan @ = x/(1-x²)^1/2

so int 1 / (1-x^2)^3/2 dx = [email protected] + C = x/(1-x²)^1/2 + C

Edit: MOCKEL WHY ARE YOU ALWAYS FASTER THAN ME??!?!
0
13 years ago
#4
Maybe you're just getting used to the new notation (e.g. 2 etc)?
0
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