_ ^ dx 2x - x^2 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }} = } $

A
${\cos ^{ - 1}}(x - 1) + c$
✓
${\sin ^{ - 1}}(x - 1) + c$
C
${\cos ^{ - 1}}(1 + x) + c$
D
${\sin ^{ - 1}}(1 - x) + c$

✓

Answer

Correct option: B.

${\sin ^{ - 1}}(x - 1) + c$

b
(b)$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {1 - {{(x - 1)}^2}} }}} = {\sin ^{ - 1}}(x - 1) + c.$

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 7.1 inderfinite integral→7.1 inderfinite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $u = \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{{x^2} + 1}}} \,dx$ and $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \,2x)} \,dx$ then

Let $f : R \to R$ be a function such that $\left| {f\left( x \right)} \right| \leq {x^2}$ , for all $x \in R$ . Then, at $x\, = 0$, $f$ is

The equation to the orthogonal trajectories of the system of parabolas $y = ax^2$ is

Let $\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{V}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ then $\frac{\text{dU}}{\text{dV}}=$

$\frac{1}{2}$
$\text{x}$
$\frac{1-\text{x}^2}{\text{x}^2-4}$
$1$

$\int {\frac{{(\sin \theta + \cos \theta )}}{{\sqrt {\sin 2\theta } }}} d\theta = $

The value of the integral $\int_{ - \pi }^\pi {{{(\cos ax - \sin bx)}^2}dx} $, ($a$ and $b$ are integer) is

$\int \limits_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$ is equal to

$\int_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}}\text{dx}$ is equal to:

$2\sqrt{2}$
$2(\sqrt{2+1})$
$0$
$2(\sqrt{2-1})$

If A satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then A^-1 exists if:

$\lambda\neq1$
$\lambda\neq2$
$\lambda\neq-1$
$\lambda\neq0$

The value of $'a' (a>0)$ for which the area bounded by the curves $y = \frac{x}{6}\, + \,\frac{1}{{{x^2}}} \, , y = 0, x = a$ and $x = 2a$ has the least value, is