Question
$\int_0^{\pi /3} {\cos 3x\,dx = } $

Answer

b
(b) Required value is $\left[ {\frac{{\sin 3x}}{3}} \right]_0^{\pi /3} = 0$.

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

The sum of the series : $(2)^2 + 2(4)^2 + 3(6)^2 + ...$ upto $10$ terms is
If $\omega $ is a complex cube root of unity, then $(1 + \omega )(1 + {\omega ^2})$ $(1 + {\omega ^4})(1 + {\omega ^8})...$to $2n$ factors =
If sum of $n$ terms of an $A.P.$ is $3{n^2} + 5n$ and ${T_m} = 164$ then $m = $
The median of $10, 14, 11, 9, 8, 12, 6$ is
The latus rectum of the parabola ${y^2} = 5x + 4y + 1$ is
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}$ be a vector such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4$. Then $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}})|^{2}$ is equal to __________ .
The random valuable $X$ follows binomial distribution $B (n, p)$ for which the difference of the mean and the variance is $1$. If $2 P(X=2)=3 P(X=1)$, then $n^2 P(X > 1)$ is equal to $......$.
The term independent of ' $x$ ' in the expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$, where $x \neq 0,1$ is equal to $.....$
Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_{1}: 3 x-4 y+12=0$, and $L _{2}: 8 x+6 y+11=0$. If $P$ lies below $L _{1}$ and above $L_{2}$, then $100(\alpha+\beta)$ is equal to
Let $\mathrm{n}$ be an odd natural number such that the variance of $1,2,3,4, \ldots, \mathrm{n}$ is $14 .$ Then $\mathrm{n}$ is equal to ..... .