Download App

STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
If $\int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx\, = \,A\,(x)\,{(\sqrt {1 - {x^2}} )^m}\, + \,C,$ for a suitable chosen integer $m$ and a function $A(x),$ where $C$ is a constant of integration, then $(A(x))^m$ equals
  • ✓
    $\frac{{ - 1}}{{27\,{x^9}}}$
  • B
    $\frac{{ - 1}}{{3\,{x^3}}}$
  • C
    $\frac{{  1}}{{27\,{x^6}}}$
  • D
    $\frac{{  1}}{{9\,{x^4}}}$
✓

Answer

Correct option: A.
$\frac{{ - 1}}{{27\,{x^9}}}$
a
$\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=\int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} d x=\int \frac{1}{x^{3}} \sqrt{\frac{1}{x^{2}}-1} d x$

${\rm{ Let }}\frac{1}{{{x^2}}} - 1 = t$

$ - \frac{2}{{{x^3}}}{\rm{d}}x = {\rm{dt}}$

${\text { Integration becomes }-\frac{1}{2} \int \sqrt{t} d t=-\frac{1}{2} \frac{(t)^{3 / 2}}{\frac{3}{2}}}$

${=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{3 / 2}}$

${=-\frac{1}{3 x^{3}}\left(1-x^{2}\right)^{3 / 2}} $

${=-\frac{1}{3 x^{3}}(\sqrt{1-x^{2}})^{3}} $

$ \Rightarrow {(A(x))^3} = {\left( { - \frac{1}{{3{x^3}}}} \right)^3} = \frac{1}{{27{x^9}}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $A =\{1,2,3,4, \ldots .10\}$ and $B =\{0,1,2,3,4\}$ The number of elements in the relation $R =\{( a , b )$ $\left.\in A \times A : 2( a - b )^2+3( a - b ) \in B \right\}$ is $.........$.
→
$A$ and $B$ are two events such that $P(A / B) =P(B/A). \neq 0$ then which of the following is correct $-$
→
The probability that a student is not a swimmer is $\frac{{1}}{{5}}$. What is the probability that out of $5$ students, $4$ are swimmers
→
The direction ratios of the line joining the points $A(4, −3, 7)$ and $B(1, 3, 5)$ are:
→
The direction ratios of the vector $3 \vec{a}+2 \vec{b}$, where $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ are
→
$\int_{ - 1}^1 {\log \frac{{2 - x}}{{2 + x}}\,dx} = $
→
If $f\left( x \right) = \frac{{2 - x\,\cos \,x}}{{2 + x\,\cos \,x}}$ and $g\left( x \right) = {\log _e}\,x$, $\left( {x > 0} \right)$ then the value of the integral $\int\limits_{\frac{{ - \pi }}{4}}^{\frac{\pi }{4}} {g\left( {f\left( x \right)} \right)} dx$ is
→
If $y=(\tan x)^{\sin x}$, then $\frac{d y}{d x}$ is equal to
→
If $f(x) = \cos \left( {{{\tan }^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}x} \right)} \right)} \right) + \sin \left( {{{\cot }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}x} \right)} \right)} \right)$ has range $\left[ {m,M),} \right.$ then number of solutions of the equation $\operatorname{sgn} \left( {\left| {x - 1} \right| - 2} \right) = \ln \left| {x - 2} \right|$ is (where sgn denotes signum function)
→
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
→