4u intergration again (1 Viewer)

[nealos]

sorry to bother again

how do u
intergrate:

(16-x^2) over x^2


using the substitution x=4sinθ

thanks
btw its from cambride 5.3 q8 or 9

 

[nealos]

i think u have taken the q differently
only the numerator is square rooted and the denominator is x^2


the answer has x's and inverse, soz i dont have the answer with me ...at skol
will post tomz

 

[Trebla]

sorry to bother again

how do u
intergrate:

(16-x^2) over x^2


using the substitution x=4sinθ

thanks
btw its from cambride 5.3 q8 or 9

x = 4sinθ
dx = 4cosθ dθ
.: ∫4cosθ sqrt(16 - 16sin²θ)dθ/16sin²θ
= ∫4cosθ sqrt(16 (1 - sin²θ))dθ/16sin²θ
= ∫4cosθ sqrt(16cos²θ)dθ/16sin²θ
= ∫4cosθ.4cosθ dθ/16sin²θ
= ∫cos²θ dθ/sin²θ
= ∫cot²θ dθ
= ∫(cosec²θ - 1) dθ
*= - cot θ - θ + c

Now, x = 4sinθ
.: sinθ = x/4
From right-angled triangle it can be deduced that:
tan θ = x/sqrt(16 - x²)
i.e. cot θ = sqrt(16 - x²)/x

Also, x = 4sinθ
.: sinθ = x/4
.: θ = sin^-1(x/4)

Hence from *
- cot θ - θ + c = - sqrt(16 - x²)/x - sin^-1(x/4) + c

I think it's right, please check my working........

 

[ianc]

thats correct

btw you can always check integrations on "The Integrator"
http://integrals.wolfram.com/index.jsp

 

[nealos]

ye thats right
but how did u intergrate cosec^2?

 

[haque]

the derivative of cot is -cosec^2 similar to tn and sec