_ ^ dx ( x + 2x) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{(\sin x + \sin 2x)}} = } $

✓
$\frac{1}{6}\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$
B
$6\log (1 - \cos x) + 2\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$
C
$6\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) + \frac{2}{3}\log (1 + 2\cos x)$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{6}\log (1 - \cos x) + \frac{1}{2}\log (1 + \cos x) - \frac{2}{3}\log (1 + 2\cos x)$

a
(a)$I = \int_{}^{} {\frac{{dx}}{{\sin x(1 + 2\cos x)}}} = \int_{}^{} {\frac{{\sin x\,dx}}{{{{\sin }^2}x(1 + 2\cos x)}}} $
$ = \int_{}^{} {\frac{{\sin x\,dx}}{{(1 - \cos x)(1 + \cos x)(1 + 2\cos x)}}} $
Now differential coefficient of $\cos x$ is $ - \sin x$ which is given in numerator and hence we make the substitution $\cos x = t \Rightarrow - \sin x\,dx = dt$
$\therefore \,\,\,I = - \int_{}^{} {\frac{{dt}}{{(1 - t)(1 + t)(1 + 2t)}}} $
We split the integrand into partial fractions
$I = - \int {\left[ {\frac{1}{{6(1 - t)}} - \frac{1}{{2(1 + t)}} + \frac{4}{{3(1 + 2t)}}} \right]} \,dt$ etc.

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 STD 12 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

For each real number $x$, let $[ x ]$ denote the greatest integer less than or equal to $x$, and let $\{ x \}= x -[ x ]$. Then the smallest positive integer $M$ for which $\int_1^M\{x\}^{[x]} d x > 1$ is

If $y = 2sin^{-1} \sqrt {(1-x)} + sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$, then $\frac{dy}{dx}$ equals-

If $\text{f(x)}=\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)}+....+\frac{\text{x}^2}{(1+\text{x}^2)}+....,$ then at $x = 0, f(x) :$

Let $\phi(\text{x})=\text{f}(\text{x})+\text{f}(2\text{a}-\text{x})$ and f'(x) > 0 for all $\text{x}\in[0,\text{a}].$ Then, $\phi(\text{x}):$

If the area (in sq. units) bounded by the parabola $y^2 =4\lambda x$ and the line $y = \lambda x$, $\lambda > 0$, is $\frac{1}{9}$, then $\lambda $ is equal to

$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}} = } $

The solution of the differential equation $\left(x^2+1\right) \frac{d y}{d x}+\left(y^2+1\right)=0$, is

A function $f$ satisfying $f ‘ \,(sin\, x)$ $= cos^2 x$ for all $x$ and $f(1) = 1$ is :

If $y = {(\sin x)^{{{(\sin x)}^{(\sin x)......\infty }}}}$, then ${{dy} \over {dx}} = $

If the system of equations

$x+y+z=2$

$2 x+4 y-z=6$

$3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then