_ ^ cose c ^2 x ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {{\rm{cose}}{{\rm{c}}^2}x\;dx} $ is equal to

A
$\cot x + c$
✓
$ - \cot x + c$
C
${\tan ^2}x + c$
D
$ - {\cot ^2}x + c$

✓

Answer

Correct option: B.

$ - \cot x + c$

b
(b)$\int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} = - \cot x + c$.

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

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

Choose the correct answer from the given four options.The identity element for the binary operation $*$ defined on $\text{Q}\sim\{0\}$ as $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}\ \forall\ \text{a, b}\in\text{Q}\sim\{0\}$ is:

Let $f(x) = 2x^3 - 3x^2 - 12x + 5$ on $[-2, 4]$. The relative maximum occurs at $x =$

The length of the perpendicular from the point $(2, -1, 4)$ on the straight line, $\frac{{x + 3}}{{10}} = \frac{{y - 2}}{{ - 7}} = \frac{z}{1}$ is

solve $\frac{{1 - \left| x \right|}}{{2 - \left| x \right|}} \ge 0$

$\int {\,\,\frac{{{x^2}\,\, + \,\,{{\cos }^2}\,x}}{{1\,\, + \,\,{x^2}}}}$ $cosec^2\, x\, dx$ is equal to :

where $‘c’$ is constant of integration .

The value of the integral $\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{dx}$ is:

If $A$ and $B$ are non $-$ zero square matrices of the same order such that $AB = 0,$ then

Choose the correct answer from the given four options.The domain of the function defined by $\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$ is:

Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is

If $X$ is brother of the son of $Y\ 's$ son. How is $X$ related to $Y?$