_ ^ dx (x - a) (x - b) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}} = } $

A
${\rm{cosec}}\,\,(a - b)\log \frac{{\sin (x - a)}}{{\sin (x - b)}} + c$
✓
${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$
C
${\rm{cosec}}(a - b)\log \frac{{\sin (x - b)}}{{\sin (x - a)}} + c$
D
${\rm{cosec}}(a - b)\log \frac{{\cos (x - b)}}{{\cos (x - a)}} + c$

✓

Answer

Correct option: B.

${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$

b
(b)$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}}} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\frac{{\sin \left\{ {(x - b) - (x - a)} \right\}}}{{\cos (x - a)\,.\,\cos (x - b)}}\,dx} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\left\{ {\frac{{\sin (x - b)}}{{\cos (x - b)}} - \frac{{\sin (x - a)}}{{\cos (x - a)}}} \right\}dx} $
$ = {\rm{cosec}}\,(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + 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

The line $x = 1, y = 2$ is:

The angle between the lines $2x = 3y = - z$ and $6x = - y = - 4z$, is ......... $^o$

Write the number of points where $f(x) = |x + 2| + |x - 3|$ is not differentiable:

If $y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)}$, then $\frac{d y}{d x}$ is equal to

The system of equation $x + y + z = 2, 3x − y + 2z = 6$ and $3x + y + z = −18$ has:

Let $f : R \to R$ be a function defined by $f(x) = - \frac{{|x{|^3} + |x|}}{{1 + {x^2}}}$; then the graph of $f(x)$ is lies in the :-

If $P(A) + P(B) = 1;$ then which of the following option explains the event $A$ and $B$ correctly?

The number which exceeds its square by the greatest possible quantity is,

If the given vectors $( - bc,\,{b^2} + bc,\,{c^2} + bc),$ $({a^2} + ac,\, - ac,\,{c^2} + ac)$ and $({a^2} + ab,\,{b^2} + ab,\, - ab)$ are coplanar, where none of $a, b $ and $ c $ is zero, then

Let $f : R \to R$ be defined by $f\left( x \right) = \frac{{\left| x \right| - 1}}{{\left| x \right| + 1}}$ then $f$ is