MCQ
$\int_{}^{} {\frac{{\cot x\tan x}}{{{{\sec }^2}x - 1}}} \;dx = $
  • A
    $\cot x - x + c$
  • B
    $ - \cot x + x + c$
  • C
    $\cot x + x + c$
  • $ - \cot x - x + c$

Answer

Correct option: D.
$ - \cot x - x + c$
d
(d) $\int_{}^{} {\frac{{\cot x\tan x}}{{{{\sec }^2}x - 1}}\,dx = \int_{}^{} {\frac{1}{{{{\tan }^2}x}}\,dx = \int_{}^{} {{{\cot }^2}x\,dx} } } $
$ = \int_{}^{} {(co{\rm{se}}{{\rm{c}}^2}x - 1)\,dx} = - \cot x - 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

Similar questions

There are$ 4$ parcels and $5$ post-offices. In how many different ways the registration of parcel can be made
Six boys and six girls sit in a row randomly. The probability that the six girls sit together
Let $y = f(x) = ax^2 + 2bx + c,$ (where $a, b, c \in  R$ and $a \neq 0)$ if $f(x) = 0$ has imaginary roots and $4a + 4b + c < 0$ then which of the following is true
The remainder on dividing $1+3+3^{2}+3^{3}+\ldots+3^{2021}$ by $50$ is
A purse contains $4$ copper coins $\& \, 3$ silver coins, the second purse contains $6$ copper coins $\& \,2$ silver coins. If a coin is drawn out of one of these purses, then the probability that it is a copper coin is :-
A solution curve of the differential equation $\left(x^2+x y+4 x+2 y+4\right) \frac{c^{\prime} y}{c^{\prime} x}-y^2=0, x>0$, passes through the point $(1,3)$. Then the solution curve

($A$) intersects $y=x+2$ exactly at one point

($B$) intersects $y=x+2$ exactly at two points

($C$) intersects $y=(x+2)^2$

($D$) does $NOT$ intersect $y=(x+3)^2$

The least value of $'a'$ for which the equation, $\frac{4}{{\sin \,x}}\,\, + \,\,\frac{1}{{1\,\, - \,\,\sin \,x}} = a$ has atleast one solution on the interval $(0, \pi /2)$ is
The partial fractions of ${{{x^2} - 5} \over {{x^2} - 3x + 2}}$ are
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $
If $A$ and $B$ are not disjoint sets, then $n(A \cup B)$ is equal to