dx x - x + 2 equals - Vidyadip

MCQ

$\int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $ equals

A
$ - \frac{1}{{\sqrt 2 }}\tan \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
B
$\frac{1}{{\sqrt 2 }}\tan \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
C
$\frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$
✓
$ - \frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$

✓

Answer

Correct option: D.

$ - \frac{1}{{\sqrt 2 }}\cot \left( {\frac{x}{2} + \frac{\pi }{8}} \right) + c$

d
(d) $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $
$ = \int {\frac{{dx}}{{\sqrt 2 (\sin x.\sin \frac{\pi }{4} - \cos x\cos \frac{\pi }{4} + 1)}}} $
$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{1 - \cos (x + \frac{\pi }{4})}}} $$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{1 - \cos 2\left( {\frac{x}{2} + \frac{\pi }{8}} \right)}}} $
$ = \frac{1}{{\sqrt 2 }}\int {\frac{{dx}}{{2{{\sin }^2}\left( {\frac{x}{2} + \frac{\pi }{8}} \right)}}} $ $ = \frac{1}{{2\sqrt 2 }}\int {{\rm{cose}}{{\rm{c}}^2}\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\,dx} $
$ = \frac{1}{{2\sqrt 2 }}\frac{{ - \cot \,\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\,}}{{12}} + c$$ = \frac{{ - 1}}{{\sqrt 2 }}\cot \,\left( {\frac{x}{2} + \frac{\pi }{8}} \right)\, + 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 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 $\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:

Let $a, b, c \in R$ be all non-zero and satisfy $a^{3}+b^{3}+c^{3}=2 .$ If the matrix $A=\left(\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right)$ satisfies $\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I},$ then a value of $abc$ can be

Let $f : R \to R$ be a function such that $f(2 - x)\, = f(2 + x)$ and $f(4 -x)\, = f(4 + x)$, for all $x \in R$ and $\int\limits_0^2 {f\left( x \right)\,dx = 5} $ . Then the value of $\int\limits_{10}^{50} {f\left( x \right)\,\,dx} $ is

Maximize Z = 4x + 6y, subject to $3\text{x}+2\text{y}\leq12,\text{x}+\text{y}\geq4,\text{x},\text{y}\geq0.$

$\int {\frac{{dx}}{{7 + 5\cos x}} = } $

If $A^2 - A + I = 0,$ then the inverse of $A$ is:

If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then:

Solution of differential equation $\frac{{dy}}{{dx}} = \frac{{y - x}}{{y + x}}$ is

Let $\mathrm{A}(-1,1)$ and $\mathrm{B}(2,3)$ be two points and $\mathrm{P}$ be a variable point above the line $A B$ such that the area of $\triangle \mathrm{PAB}$ is $10$ . If the locus of $\mathrm{P}$ is $\mathrm{ax}+\mathrm{by}=15$, then $5 a+2 b$ is :

Let $f(x) = \left| {{\mkern 1mu} \begin{array}{*{20}{c}}
{{x^3}}&{\sin x}&{\cos x} \\
6&{ - 1}&0 \\
p&{{p^2}}&{{p^3}}
\end{array}{\mkern 1mu} } \right|$, where $p$ is a constant. Then ${{{d^3}} \over {d{x^3}}}\left\{ {f(x)} \right\}$ at $x = 0$ is