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Question: How do you integrate \(\int{\dfrac{{{x}^{2}}}{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}...

How do you integrate ∫x2(a2−x2)32dx\int{\dfrac{{{x}^{2}}}{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}}dx by trigonometric substitution?

Explanation

Solution

We are given a function that we need to integrate with respect to xx. Since, it is a quadratic function; we know that it would require some substitution. We are asked to do this by trigonometric substitution for which we would use various trigonometric formulae and do the substitution accordingly.

Complete step-by-step solution:
Since we have a2−x2a^2-x^2 in the denominator we can think of some substitution which would be a multiple of aa times the cosine or the sine function, because then we could take the a2a^2 out in common and the term left behind will be of the type of 1−cos(x)1-cos\left( x \right) or 1−sin2x1-sin^2x for which we would use the identity:
sin2x+cos2x=1sin^2x+cos^2x=1
So, let’s substitute:
x=a×sin(t)x=a\times sin\left(t\right)
  ⟹  dx=a×cos(t)\implies dx=a\times cos\left(t\right)
We have:
∫x2(a2−x2)32dx=∫a2sin2t(a2−a2sin⁡2t)32a(cos(t))dt\int{\dfrac{{{x}^{2}}}{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx}=\int{\dfrac{{{a}^{2}}si{{n}^{2}}t}{{{\left( {{a}^{2}}-{{a}^{2}}{{\sin }^{2}}t \right)}^{\dfrac{3}{2}}}}}a\left( cos\left( t \right) \right)dt
=∫a2sin2t(a2(1−sin⁡2t))32a(cos(t))dt=\int{\dfrac{{{a}^{2}}si{{n}^{2}}t}{{{\left( {{a}^{2}}\left( 1-{{\sin }^{2}}t \right) \right)}^{\dfrac{3}{2}}}}}a\left( cos\left( t \right) \right)dt
=∫a2sin2t(a2(cos⁡2t))32a(cos(t))dt=\int{\dfrac{{{a}^{2}}si{{n}^{2}}t}{{{\left( {{a}^{2}}\left( {{\cos }^{2}}t \right) \right)}^{\dfrac{3}{2}}}}}a\left( cos\left( t \right) \right)dt
Cancelling out the power of 2 and cubing the function in the denominator, we get:
Now, we use the formula:
sinθcosθ=tanθ\dfrac{sin\theta}{cos\theta}=tan\theta
Using this, we get:
=∫sin2tcos2tdt=\int \dfrac{sin^2t}{cos^2t}dt
=∫tan2tdt=\int tan^2tdt
Now, we use:
tan2θ=sec2θ−1ta{{n}^{2}}\theta =se{{c}^{2}}\theta -1
∫tan⁡2tdt=∫(sec2t−1)dt\int{{{\tan }^{2}}tdt}=\int{\left( se{{c}^{2}}t-1 \right)}dt
=tan⁡t−t+c=\tan t-t+c
We have used the formula ∫sec⁡2θdθ=tanθ\int{{{\sec }^{2}}\theta }d\theta =tan\theta
=sin⁡tcos⁡t−t+c=\dfrac{\sin t}{\cos t}-t+c
Now, use the identity cos2x+sin2x=1cos^2x+sin^2x=1
=sin⁡t1−sin⁡2t−t+c=\dfrac{\sin t}{\sqrt{1-{{\sin }^{2}}t}}-t+c
Now, since
x=a×sin⁡tx=a\times \sin t
⇒sin⁡t=xa\Rightarrow \sin t=\dfrac{x}{a}
And t=sin−1xat=sin^{-1}\dfrac{x}{a}
Hence, the integral becomes:
∫x2(a2−x2)32dx=xa1−(xa)2−sin−1(xa)+c\int{\dfrac{{{x}^{2}}}{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}}dx=\dfrac{\dfrac{x}{a}}{\sqrt{1-{{\left( \dfrac{x}{a} \right)}^{2}}}}-si{{n}^{-1}}\left( \dfrac{x}{a} \right)+c
=xa2−x2−sin−1(xa)+c=\dfrac{x}{\sqrt{a^2-x^2}}- sin^{-1}\left(\dfrac{x}{a}\right)+c
Hence, the integral has been found.

Note: We could have also used the substitution of sine but then it would have turned into a cotangent function. Since, we do more problems involving tangent, the chances of mistakes in the formulae will be lesser in this case. Moreover, to check whether the result is correct or not you could easily differentiate the obtained function and check whether it matches with the function given in the question itself. Also, you need to be aware about all the formulae of integration so that you don’t get stuck and you don’t perform wrong operations by mistake.