^ _0 1-x 1+x dx= - Vidyadip

MCQ

$\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=$

✓
$\sqrt{1-\pi^2}-1$
B
$\frac{\pi}{2}-1$
C
$\frac{\pi}{2}+1$
D
${\pi}+{1}$

✓

Answer

Correct option: A.

$\sqrt{1-\pi^2}-1$

We have,

$\text{I}=\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$

$=\int\limits^\pi_0\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{\sqrt{1-\text{x}}}{\sqrt{1-\text{x}}}\Big]\text{dx}$

$=\int\limits^\pi_0\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$

$=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}-\int\limits^\pi_0\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$

Putting $1 -\text{x}^2=\text{t}$

$\Rightarrow - 2\text{x}\text{ dx}=\text{dt}$

$\Rightarrow \text{x}\text{ dx}=\frac{-\text{dt}}{2}$

when $\text{x}\rightarrow0;\text{t}\rightarrow1$

and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-\pi^2$

$\therefore\ \text{I}=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}-\int\limits^{(1-\pi^2)}_1\frac{-\text{dt}}{2\sqrt{\text{t}}}$

$=\big[\sin^{-1}\text{x}\big]^{\pi}_0+\frac{2}{2}\big[\sqrt{\text{t}}\big]^{1-\pi^2}_1$

$=\big[0-0\big]+\Big[\sqrt{1-\pi^2}-\sqrt{1}\Big]$

$=\sqrt{1-\pi^2}-1$

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 INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The value of $m$ for which the function $f(x) = \left\{ \begin{array}{l}m{x^2},\,x \le 1\\\,\,\,\,2x,\,x > 1\end{array} \right.$ is differentiable at $x = 1$, is

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|$is

Let $[x]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}$ is $(-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}$, then the value of $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is:

$\int_{}^{} {x{{\sin }^2}x\;dx = } $

If $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and ${A_1},{B_1},{C_1}$denote the co-factors of ${a_1},{b_1},{c_1}$ respectively, then the value of the determinant $\left| {\begin{array}{*{20}{c}}{{A_1}}&{{B_1}}&{{C_1}}\\{{A_2}}&{{B_2}}&{{C_2}}\\{{A_3}}&{{B_3}}&{{C_3}}\end{array}} \right|$ is

$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $

The function $f(x)=k x$ - $\sin x$ is strictly increasing for

Let $A = \{1, 2, 3, …. n\}$ and $B = \{a, b\}$. Then the number of surjections from $A$ into $B$ is:

Let the solution curve $y=y(x)$ of the differential equation $\left(1+ e ^{2 x }\right)\left(\frac{ dy }{ dx }+ y \right)=1$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim _{x \rightarrow \infty} e ^{x} y(x)$ is equal to.

The mean and variance of a random variable $X$ having a binomial distribution are $4$ and $2$ respectively, then $P(X = 1)$ is