I know the answer is (-sqrt(324-x^2))/(324x) but when I try to solve it I get (-arcsin(x/18))/(x). I am using the method of trigonometric substitution and I set x^2/324=sin^2\theta then x=18*sin\theta and dx=18*cos\theta d\theta. Please help me understand this.
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Let x = √324 sin t = 18 sin t.
==> dx = 18 cos t dt.
So, ∫ dx / [x^2 √(324 - x^2)]
= ∫ 18 cos t dt / [(18 sin t)^2 √(324 cos^2(t))]
= ∫ dt/(18 sin t)^2
= (1/324) ∫ csc^2(t) dt
= (1/324) * -cot t + C
= (-1/324) * √(324 - x^2)/x + C, via sin t = x/18 and 'sohcahtoa'
I hope this helps!
==> dx = 18 cos t dt.
So, ∫ dx / [x^2 √(324 - x^2)]
= ∫ 18 cos t dt / [(18 sin t)^2 √(324 cos^2(t))]
= ∫ dt/(18 sin t)^2
= (1/324) ∫ csc^2(t) dt
= (1/324) * -cot t + C
= (-1/324) * √(324 - x^2)/x + C, via sin t = x/18 and 'sohcahtoa'
I hope this helps!