Download App

Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{\pi / 6}^{\pi / 4} \operatorname{cosec} 2 x d x=$
  • A
    $\log 3$
  • B
    $\log \sqrt{3}$
  • C
    $\log 9$
  • ✓
    $\frac{1}{2} \log \sqrt{3}$
✓

Answer

Correct option: D.
$\frac{1}{2} \log \sqrt{3}$
(D)
$\int_{\pi / 6}^{\pi / 4} \operatorname{cosec} 2 x d x=\frac{1}{2}[\log \tan x]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$
$=\frac{1}{2}\left[\log \tan \frac{\pi}{4}-\log \tan \frac{\pi}{6}\right]$
$=\frac{1}{2} \log \sqrt{3}$

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 "MCQ" from Definite Integration→Definite Integration — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}-\text{Ky}=0, \text{y}(0)=1$ approaches to zero when $\text{x}\rightarrow\propto$ if,
→
The length of the perpendicular drawn from the point $(1,2,3)$ to the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is
→
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
→
General solution of the equation $\cot \theta-\tan \theta=2$ is
→
The coordinates of the point on the ellipse $16x^2+ 9y^2= 400$ where the ordinate decreases at the same rate at which the abscissa increases, are:
→
Let $\text{A}=\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix},$ where $0\leq\theta\leq2\pi.$ Then:
→
If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$, then $\frac{|\overline{ a } \times \overline{ b }|}{|\overline{ a } \cdot \overline{ b }|}$ is
→

If $f(x)=\left\{\begin{array}{ccc}3\left(1-2 x^2\right) & ; & 0<x<1 \\ 0 & ; & \text { otherwise }\end{array}\right.$
is a probability density function of $X$, then $P\left(\frac{1}{4}<x<\frac{1}{3}\right)$ is

→
$\int\frac{\sin^6\text{x}}{\cos^8\text{x}}\text{ dx}=$
→
The Boolean Expression $(p \wedge \sim q) \vee q \vee(\sim p \wedge q)$ is equivalent to :
→