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Indefinite Integrals question

2019 · 11 Jan · Shift 1 · Q32
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Indefinite Integrals question

2019 · 11 Jan · Shift 1 · Q32

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}∫x41−x2​​ dx = A(x) (1−x2)m{\left( {\sqrt {1 - {x^2}} } \right)^m}(1−x2​)m + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
  1. A
    127x6{1 \over {27{x^6}}}27x61​
  2. B
    −127x9{{ - 1} \over {27{x^9}}}27x9−1​
  3. C
    19x4{1 \over {9{x^4}}}9x41​
  4. D
    13x3{1 \over {3{x^3}}}3x31​
View written solutionFree

Correct answer: B

  1. Given integral

We need to evaluate

I=∫1−x2x4 dxI=\int \frac{\sqrt{1-x^2}}{x^4}\,dxI=∫x41−x2​​dx

and write it in the form

I=A(x) (1−x2)m+CI=A(x)\,\big(\sqrt{1-x^2}\big)^m + CI=A(x)(1−x2​)m+C

for some integer mmm and function A(x)A(x)A(x).

We are then asked to find (A(x))m(A(x))^m(A(x))m.

  1. Try a form suggested by differentiation

Because the integrand contains 1−x2\sqrt{1-x^2}1−x2​ and powers of xxx, let us try differentiating

(1−x2)3/2x3.\frac{(1-x^2)^{3/2}}{x^3}.x3(1−x2)3/2​.

Let

F(x)=(1−x2)3/2x3=(1−x2)3/2x−3.F(x)=\frac{(1-x^2)^{3/2}}{x^3}=(1-x^2)^{3/2}x^{-3}.F(x)=x3(1−x2)3/2​=(1−x2)3/2x−3.

Then

F′(x)=x−3⋅ddx(1−x2)3/2+(1−x2)3/2⋅ddx(x−3).F'(x)=x^{-3}\cdot \frac{d}{dx}(1-x^2)^{3/2}+(1-x^2)^{3/2}\cdot \frac{d}{dx}(x^{-3}).F′(x)=x−3⋅dxd​(1−x2)3/2+(1−x2)3/2⋅dxd​(x−3).

Now,

ddx(1−x2)3/2=32(1−x2)1/2(−2x)=−3x1−x2,\frac{d}{dx}(1-x^2)^{3/2}=\frac{3}{2}(1-x^2)^{1/2}(-2x)=-3x\sqrt{1-x^2},dxd​(1−x2)3/2=23​(1−x2)1/2(−2x)=−3x1−x2​,

and

ddx(x−3)=−3x−4.\frac{d}{dx}(x^{-3})=-3x^{-4}.dxd​(x−3)=−3x−4.

So,

F′(x)=x−3(−3x1−x2)+(1−x2)3/2(−3x−4).F'(x)=x^{-3}(-3x\sqrt{1-x^2})+(1-x^2)^{3/2}(-3x^{-4}).F′(x)=x−3(−3x1−x2​)+(1−x2)3/2(−3x−4).

Simplify:

F′(x)=−3x−21−x2−3x−4(1−x2)3/2.F'(x)=-3x^{-2}\sqrt{1-x^2}-3x^{-4}(1-x^2)^{3/2}.F′(x)=−3x−21−x2​−3x−4(1−x2)3/2.

Factor −31−x2-3\sqrt{1-x^2}−31−x2​:

F′(x)=−31−x2(1x2+1−x2x4).F'(x)=-3\sqrt{1-x^2}\left(\frac{1}{x^2}+\frac{1-x^2}{x^4}\right).F′(x)=−31−x2​(x21​+x41−x2​).

Inside the bracket,

1x2+1−x2x4=x2+1−x2x4=1x4.\frac{1}{x^2}+\frac{1-x^2}{x^4}=\frac{x^2+1-x^2}{x^4}=\frac{1}{x^4}.x21​+x41−x2​=x4x2+1−x2​=x41​.

Hence,

F′(x)=−31−x2x4.F'(x)=-3\frac{\sqrt{1-x^2}}{x^4}.F′(x)=−3x41−x2​​.

Therefore,

1−x2x4=−13F′(x).\frac{\sqrt{1-x^2}}{x^4}=-\frac13 F'(x).x41−x2​​=−31​F′(x).

So,

I=−13⋅(1−x2)3/2x3+C.I=-\frac13\cdot \frac{(1-x^2)^{3/2}}{x^3}+C.I=−31​⋅x3(1−x2)3/2​+C.
  1. Match with the required form

Since

(1−x2)3/2=(1−x2)3,(1-x^2)^{3/2}=\big(\sqrt{1-x^2}\big)^3,(1−x2)3/2=(1−x2​)3,

we can write

I=(−13x3)(1−x2)3+C.I=\left(-\frac{1}{3x^3}\right)\big(\sqrt{1-x^2}\big)^3 + C.I=(−3x31​)(1−x2​)3+C.

Thus,

m=3,A(x)=−13x3.m=3, \qquad A(x)=-\frac{1}{3x^3}.m=3,A(x)=−3x31​.
  1. Compute (A(x))m(A(x))^m(A(x))m

We need

(A(x))m=(−13x3)3=−127x9.(A(x))^m=\left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}.(A(x))m=(−3x31​)3=−27x91​.
  1. Compare with options

The value is

−127x9,-\frac{1}{27x^9},−27x91​,

which is Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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