_ ^ cosec - cosec + ;d = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{\rm{cosec}}\theta - \cot \theta }}{{{\rm{cosec}}\theta + \cot \theta }}} \;d\theta = $

✓
$2{\rm{cosec}}\theta - 2\cot \theta - \theta + c$
B
$2\,{\rm{cosec}}\theta - 2\cot \theta + \theta + c$
C
$2\,{\rm{cosec}}\theta + 2\cot \theta - \theta + c$
D
None of these

✓

Answer

Correct option: A.

$2{\rm{cosec}}\theta - 2\cot \theta - \theta + c$

a
(a)$\int_{}^{} {\frac{{{\rm{cosec}}\theta - \cot \theta }}{{{\rm{cosec}}\theta + \cot \theta }}\,d\theta } = \int_{}^{} {{{({\rm{cosec}}\theta - \cot \theta )}^2}d\theta } $
$ = \int_{}^{} {{\rm{cose}}{{\rm{c}}^2}\theta \,d\theta } + \int_{}^{} {{{\cot }^2}\theta \,d\theta } - 2\int_{}^{} {{\rm{cosec}}\theta \cot \theta \,d\theta } $
$ = \int_{}^{} {(2{\rm{cose}}{{\rm{c}}^2}\theta - 1)\,d\theta } - 2\int_{}^{} {{\rm{cosec}}\theta \cot \theta \,d\theta } $
$ = 2{\rm{cosec}}\theta - 2\cot \theta - \theta + 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

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ equals to

Choose the correct answer from the given four options. The angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and $4,$ respectively, and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ is:

If $\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix},$ then $A^n =$

How many lines through the origin in make equal angles with the coordinate axis:

$(i)$ $f (x)$ is continuous and defined for all real numbers

$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4) = 0$

$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$

$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.

$(v)$ the signs of $f '(x)$ is given below

On the possible graph of $y = f (x)$ we have

If $A,\,B,\,C$ are the vertices of a triangle whose position vectors are $a, b, c $ and $G$ is the centroid of the $\Delta ABC,$ then $\overrightarrow {GA} + \overrightarrow {GB} \, + \overrightarrow {GC} $ is

If $A$ is an invertible matrix, then det $(A^{-1})$ is equal to:

Let $A(3, 0, -1), B(2, 10, 6)$ and $C(1, 2, 1)$ be the vertices of a triangle and $M$ be the midpoint of $AC$. If $G$ divides $BM$ in the ratio, $2 : 1$, then $\cos \,\left( {\angle GOA} \right)$ ($O$ being he origin) is equal to

$\int {\frac{{{{(x + 1)}^2}\,\,dx}}{{x({x^2} + 1)}}} $ is equal to

If $\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2},$ then $(1-\text{x})^2\text{f}\ ''(\text{x})-\text{xf}(\text{x})=$