dx 1 - x - x = - Vidyadip

MCQ

$\int {\frac{{dx}}{{1 - \cos x - \sin x}}} = $

A
$\log |1 + \cot x/2| + c$
B
$\log |1 - \tan x/2| + c$
✓
$\log |1 - \cot x/2| + c$
D
$\log |1 + \tan x/2| + c$

✓

Answer

Correct option: C.

$\log |1 - \cot x/2| + c$

c
(c)$I = \int {\frac{{dx}}{{1 - \cos x - \sin x}}} $
$ = \int {\frac{{dx}}{{1 - \frac{{[(1 - {{\tan }^2}(x/2)]}}{{[(1 + {{\tan }^2}(x/2)]}} - \frac{{2\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}}}}} $
$ = \int {\frac{{{{\sec }^2}(x/2).dx}}{{1 + {{\tan }^2}(x/2) - 1 + {{\tan }^2}(x/2) - 2\tan (x/2)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{2{{\tan }^2}(x/2) - 2\tan (x/2)}}} $ $ = \int {\frac{{\frac{1}{2}.{{\sec }^2}\left( {\frac{x}{2}} \right)\,dx}}{{{{\tan }^2}\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right)}}} $
Put $\tan (x/2) = t$==> $\frac{1}{2}{\sec ^2}(x/2)\,dx = dt$
$I = \int {\frac{{dt}}{{{t^2} - t}}} $ $ = \int {\frac{{dt}}{{t\,(t - 1)}}} $$ = \int {\left[ {\frac{1}{{t - 1}} - \frac{1}{t}} \right]\,dt} $ ( Put tan $x = t,$ $\therefore {\sec ^2}x{\rm{ }}dx = dt$)
$ = \int {\frac{{dt}}{{t - 1}} - \int {\frac{{dt}}{t}} } = \log (t - 1) - \log t + c = \log \left| {\frac{{t - 1}}{t}} \right| + c$
$ = \log \left| {\,\frac{{\tan (x/2) - 1}}{{\tan (x/2)}}\,} \right| + c$$ = \log \left| {\,1 - \cot \,\frac{x}{2}\,} \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 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

The area bounded by the parabola $x^2=4 y$ and its latus rectum is:

For positive integer $n$, define

$f(n)=n+\frac{16+5 n-3 n^2}{4 n+3 n^2}+\frac{32+n-3 n^2}{8 n+3 n^2}+\frac{48-3 n-3 n^2}{12 n+3 n^2}+\ldots+\frac{25 n-7 n^2}{7 n^2}$

Then, the value of $\lim _{ n \rightarrow \infty} f( n )$ is equal to

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:

$\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
$\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
$\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
$\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$

If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$

$\frac{1}{5}$
$\frac{3}{10}$
$\frac{1}{2}$
$\frac{1}{2}$

Vikas printing company takes fee of Rs. 28 to print a oversized poster and Rs. 7 for each colour of ink used. Raaj printing company does the same work and charges poster for Rs. 34 and Rs. 5.50 for each colour of ink used. If z is the colours of ink used, find the values of z such that Vikas printing company would charge more to print a poster than Raaj printing company.

$\text{z}<4$
$2\leq\text{z}\leq4$
$4\leq\text{z}\leq7$
$\text{z}>4$

The number of solution of the following equations ${x_2} - {x_3} = 1,\,\, - {x_1} + 2{x_3} = - 2,$ ${x_1} - 2{x_2} = 3$ is

Let $a=i-k,\,\,\,b=xi+j+(1-x)\,k$ ,$c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

Let $y=y(x)$ be the solution of the differential equation, $x y^{\prime}-y=x^{2}(x \cos x+\sin x), x>0$ If $y (\pi)=\pi,$ then $y ^{\prime \prime}\left(\frac{\pi}{2}\right)+ y \left(\frac{\pi}{2}\right)$ is equal to

The adjoint matrix of $\left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]$is