_ ^ e^ - x cose c ^2 (2 e^ - x + 5) ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^{ - x}}{\rm{cose}}{{\rm{c}}^2}(2{e^{ - x}} + 5)} \;dx = $

✓
$\frac{1}{2}\cot (2{e^{ - x}} + 5) + c$
B
$ - \frac{1}{2}\cot (2{e^{ - x}} + 5) + c$
C
$2\cot (2{e^{ - x}} + 5) + c$
D
$ - 2\cot (2{e^{ - x}} + 5) + c$

✓

Answer

Correct option: A.

$\frac{1}{2}\cot (2{e^{ - x}} + 5) + c$

a
(a) Put $2{e^{ - x}} + 5 = t \Rightarrow - 2{e^{ - x}}dx = dt,$ then
$\int_{}^{} {{e^{ - x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}(2{e^{ - x}} + 5)\,dx} = - \frac{1}{2}\int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}t\,dt} $
$ = \frac{1}{2}\cot t = \frac{1}{2}\cot (2{e^{ - x}} + 5) + 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

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}=$

$\pi\ln 2$
$-\pi\ln2$
$0$
$-\frac{\pi}{2}\ln2$

Let $f: R \rightarrow R$ be a differentiable function that satisfies the relation $f ( x + y )= f ( x )+ f ( y )-1, \forall x$, $y \in R$. If $f ^{\prime}(0)=2$, then $|f(-2)|$ is equal to $.........$.

If A and B are square matrices of the same order, then (A + B)(A - B) is equal to:

A² - B²
A² - BA - AB - B²
A² - B² + BA - AB
A² - BA + B²+ AB

If the curve $y=y(x)$ is the solution of the differential equation $2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x > 0$ which passes through the point $\left(1,1-\frac{4}{3} \log _{e} 2\right),$ then the value of $y(16)$ is equal to :

If f : A → B is surjective then:

no two elements of A have the same image in B
every element of A has an image in B
every element of B has at least one pre-image in A
A and B are finite non empty sets

Find the relationship be $a$ and $b$ so that the function $f$ defined by $f(x) = \left\{ {\begin{array}{*{20}{l}}
{ax + 1,{\rm{ if }}\,x\, \le \,3}\\
{bx + 3,{\rm{ if }}\,x\, > \,3}
\end{array}} \right.$ is continous at $x=3.$

A random variable has the following probability distribution:

X = x_i	0	1	2	3	4	5	6	7
P(X = X_i)	0	2p	2p	3p	p²	2p²	7p²	2p

$\frac{1}{10}$
$-1$
$-\frac{1}{10}$
$\frac{1}{5}$

If the line $\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$, then $4 \lambda+9 \mu$ is equal to :

The corner points of the bounded feasible region are $(0,0),(2,0),(4,2),(2,4)$ and $\left(0, \frac{10}{3}\right)$

Then for the objective function $z=-x+2 y$

$(i)$ Maximum value of $z$ has at $\ldots \ldots \ldots . . .$

$(ii)$ Minimum value of $z$ has at $\ldots \ldots \ldots . . .$

$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots . . .$

$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots . . .$

Choose the correct option from given four options:
If $\frac{\text{x}^3\text{dx}}{\sqrt{1+\text{x}^2}}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then:

$\text{a}=\frac{1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=-1$
$\text{a}=\frac{1}{3},\text{b}=-1$