Evaluate: x 2 x 3 x d x - Vidyadip

Question

Evaluate: $\int \tan \ x \tan 2 x \tan \ 3 x\ d x$

✓

Answer

$(d) $ : Let $I=\int \tan\ x\ \tan\ 2 x \tan \ 3 x\ d x$
Since $, \tan 3 x=\tan (2 x+x)=\frac{\tan \ 2 x+\tan x}{1-\tan x \tan \ 2 x}$
$\Rightarrow \tan \ x \tan 2 x \tan \ 3 x=\tan \ 3 x-\tan \ 2 x-\tan x ...(i)$
$\therefore I=\int(\tan 3 x-\tan 2 x-\tan x) d x \ ($From $(i))$
$=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$

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 Integrals→Integrals — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Matrices A and B will be inverse of each other only if:

AB = BA
AB = BA = 0
AB = 0, BA = I
AB = BA = I

The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80, x, y ≥ 0 is:

320
300
230
none of these

The direction cosines of the ray P(1, -2, 4) and Q(-1, 1, -2) are:

$\big(-2, -3, -6\big)$
$\big(2, -3, -6\big)$
$\Big(\frac{2}{7},\frac{3}{7},\frac{6}{7}\Big)$
$\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

The element in the second row and third column of the matrix $\begin{bmatrix}4&\text{amp; }5&\text{amp; }6 \\3 &\text{amp;}-4&\text{amp; }3\\2 &\text{amp; }1&\text{amp; }0 \end{bmatrix}$ is$:$

The value of $\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)$ is:

$\frac{5}{17}$
$\frac{6}{17}$
$\frac{3}{17}$
$\frac{4}{17}$

Let $\vec{a}$ and $\vec{b}$ are unit vectors enclosing an angle $\theta$ and $|\vec{a}+\vec{b}| < 1$. Which of the following is true?
$(i) \theta=\frac{\pi}{2}$
$(ii) \theta < \frac{\pi}{3}$
$(iii) \pi \geq \theta > \frac{2 \pi}{3}$
$(iv) \cos \theta < -\frac{1}{2}$

A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3) are the vertices of a tringle ABC. if the bisector of $\angle\text{ABC}$ meets BC at D, then coordinates of D are:

$\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
$\Big(-\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
$\Big(\frac{19}{8},-\frac{57}{16},\frac{17}{16}\Big)$
$\text{none of these}$

Using determinants, find the area $($in sq. units$)$ of triangle with vertices $(-3,5),(3,-6)$ and $(7,2)$.

The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:

If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of k is:

3
4
6
7