( x + x ) is equal to- - Vidyadip

MCQ

$\int {(\sqrt {\tan x} + \sqrt {\cot x} )} $ is equal to-

A
$sin^{-1} (sinx -cosx) + C$
✓
$\sqrt 2 \,sin^{-1} (sinx -cosx) + C$
C
$\sqrt 2 cos^{-1} (sinx -cosx) + C$
D
none of these

✓

Answer

Correct option: B.

$\sqrt 2 \,sin^{-1} (sinx -cosx) + C$

b
$\int \sqrt{\tan x}+\sqrt{\cot x} d x$

$\int \frac{(\sin x+\cos x) d x}{\sqrt{\sin x \cos x}}$

$\sqrt{2} \int \frac{(\sin x+\cos x) d x}{\sqrt{1-(1-\sin 2 x)}}$

$\sqrt{2} \int \frac{(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}}$

$\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c$

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

If $\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\4&0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4\\2&{ - 5}\end{array}} \right]$, then $(a,b,c,d) = $

If $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{1 + {a^3}}\\b&{{b^2}}&{1 + {b^3}}\\c&{{c^2}}&{1 + {c^3}}\end{array}\,} \right| = 0$ and $a = (1,\,a,\,{a^2}),\,b = (1,\,b,\,{b^2}),$ and $c = (1,\,c,\,{c^2})$ are non-coplanar vectors, then $abc$ is equal to

If $A =\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]$, then $|\operatorname{adj}(\operatorname{adj}(2 A ))|$ is equal to:

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ - 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} - 2{x^2} - \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then

Let $f(x)=\int_0^x g(t) \log _e\left(\frac{1-t}{1+t}\right) d t$, where $g$ is a continuous odd function. If $\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$, then $\alpha$ is equal to..............

Let $y=y(x)$ be the solution of the differential equation $x^3 d y+(x y-1) d x=0, x>0$, $y\left(\frac{1}{2}\right)=3-e$. Then $y(1)$ is equal to

If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to

If $y = a\cos \left( {\ln x} \right) + b\sin \left( {\ln x} \right)$, then ${x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}$ is equal to

If function $f(x)=\left\{\begin{array}{c}\frac{x^2-16}{x-4}, x \neq 4 \\ k, x=4\end{array}\right.$, is continuous at $x=4$, then value of $k$ is :

$\mathop {Limit}\limits_{h\,\, \to \,\,0} \frac{{\int\limits_a^{x\, + \,h} {\,\ell {n^2}t\,\,\,dt} \,\, - \,\,\int\limits_a^x {\,\ell {n^2}t\,\,\,dt} }}{h}$ =