_ ^ 2x 5x 3x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}} \;dx = $

A
$\log \sin 3x - \log \sin 5x + c$
B
$\frac{1}{3}\log \sin 3x + \frac{1}{5}\log \sin 5x + c$
✓
$\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$
D
$3\log \sin 3x - 5\log \sin 5x + c$

✓

Answer

Correct option: C.

$\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$

c
(c)$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}\,dx} = \int_{}^{} {\frac{{\sin (5x - 3x)}}{{\sin 5x\sin 3x}}\,dx} $
$ = \int_{}^{} {\frac{{\sin 5x\cos 3x - \cos 5x\sin 3x}}{{\sin 5x\sin 3x}}\,dx} $
$ = \frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + 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

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on set $A = \{1, 2, 3\}$. Then $RoS =$

If $A = \left( {\begin{array}{*{20}{c}}i&1\\0&i\end{array}} \right)$, then ${A^4}$ equals

Which of the following sets are convex?

$\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\geq1\}$
$\{(\text{x},\text{y}):\text{y}^2\geq\text{x}\}$
$\{(\text{x},\text{y}):3\text{x}^2+4\text{y}^2\geq5\}$
$\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$

$\text{f}(\text{x})=1+2\sin\text{x}+3\cos^{2}\text{x}, 0<\text{x}<\frac{2\pi}{3}$ is :

Minimum at $\text{x}=\frac{\pi}{2}$
Maximum at $\text{x}=\sin^{-1}(\frac{1}{\sqrt{3}})$
Minimum at $\text{x}=\frac{\pi}{6}$
Maximum at $\sin^{-1}(\frac{1}{6})$

$\smallint \frac{{dx}}{{\cos x + \sqrt 3 \sin x}} = $

If the vector $\vec a=i-j+2k ,\vec b=2i+4j+k $ and $\vec c=$ $\alpha i+j+\beta k$ are mutually orthogonal then $(\alpha ,\beta ) =$

Area between the curves $y = x^3$ and $y = \sqrt x$ is

If for $x \geq 0, y=y(x)$ is the solution of the differential equation $(\mathrm{x}+1) \mathrm{d} \mathrm{y}=\left((\mathrm{x}+1)^{2}+\mathrm{y}-3\right) \mathrm{d} \mathrm{x}, \mathrm{y}(2)=0$ then $y(3)$ is equal to

If ${I_1} = \int_e^{{e^2}} {\frac{{dx}}{{\log x}}} $ and ${I_2} = \int_1^2 {\frac{{{e^x}}}{x}\,dx,} $ then

The differention equation $\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}},\text{n}>2$ can be reduced to linear from by substituting:

$\text{z}=\text{y}^{\text{n}-1}$
$\text{z}=\text{y}^{\text{n}}$
$\text{z}=\text{y}^{\text{n}+1}$
$\text{z}=\text{y}^{1-\text{n}}$