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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Indefinite Integration

If ∫ 1-x^ 2 x^ 4 d x=A(x) ( 1-x^ 2 )^ m +C, for a suitable chosen...

If ∫1−x2x4dx=A(x)\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,forasuitablechosenintegerfor a suitable chosen integer\mathrm{m} and a function A(x)\mathrm{A}(\mathrm{x}), where C\mathrm{C} is a constant of integration, then (A(x))m(\mathrm{A}(\mathrm{x}))^{\mathrm{m}} equals :

Options

  1. A−127x9\frac{-1}{27 x^{9}}
  2. B−13x3\frac{-1}{3 x^{3}}
  3. C127x6\frac{1}{27 x^{6}}
  4. D19x4\frac{1}{9 x^{4}}

Correct answer

A. −127x9\frac{-1}{27 x^{9}}

Step-by-step solution

A(x)(1−x2)m+C=∫1−x2x4dxA(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x =∫1x2−1x3dx=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d x Let 1x2−1=u2\frac{1}{x^{2}}-1=u^{2} ⇒−2x3=2ududx\Rightarrow-\frac{2}{x^{3}}=\frac{2 u d u}{d x} dxx3=−udu\frac{d x}{x^{3}}=-u d u A(x)(1−x2)m+C=∫(−u2)du=−u33+CA(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int\left(-u^{2}\right) d u=-\frac{u^{3}}{3}+C =−13(1x2−1)32+C=−13⋅1x3⋅(1−x2)32+C=−13x3(1−x2)3+C\begin{array}{l} =-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \\ =-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \\ =\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C \end{array} Compare both sides, ⇒A(x)=−13x3 and m=3⇒(A(x))3=−127x9\begin{array}{l} \Rightarrow A(x)=-\frac{1}{3 x^{3}} \text{ and } m=3 \\ \Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}} \end{array}
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