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

MCQ

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

A
$\frac{{\sin x + \cos x}}{{x\sin x + \cos x}}$
B
$\frac{{x\sin x - \cos x}}{{x\sin x + \cos x}}$
✓
$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$
D
None of these

✓

Answer

Correct option: C.

$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$

c
(c) Differentiation of$x\sin x + \cos x$is $x\cos x,$ then
$I = \int_{}^{} {\frac{{{x^2}dx}}{{{{(x\sin x + \cos x)}^2}}}} = \int_{}^{} {\frac{{x\cos x}}{{{{(x\sin x + \cos x)}^2}}}.\frac{x}{{\cos x}}dx} $
Integrate by parts $\left[ {\int_{}^{} {\frac{1}{{{t^2}}}\,dt = - \frac{1}{t}} } \right]$
$\therefore \,\,\,I = \frac{{ - 1}}{{(x\sin x + \cos x)}}.\frac{x}{{\cos x}}$
$ + \int_{}^{} {\frac{1}{{(x\sin x + \cos x)}}} .\frac{{\cos x\,.\,1 - x( - \sin x)}}{{{{\cos }^2}x}}\,dx$
$ = - \frac{1}{{x\sin x + \cos x}}.\frac{x}{{\cos x}} + \int_{}^{} {{{\sec }^2}x\,dx} $
$ = - \frac{1}{{x\sin x + \cos x}}.\frac{x}{{\cos x}} + \frac{{\sin x}}{{\cos x}}$
$ = \frac{{ - x + x{{\sin }^2}x + \sin x\cos x}}{{(x\sin x + \cos x)\cos x}}$
$ = \frac{{\sin x\cos x - x(1 - {{\sin }^2}x)}}{{(x\sin x + \cos x)\cos x}}$$ = \frac{{\sin x - x\cos x}}{{x\sin x + \cos x}}$.

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

If $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$, then $k = $

Let $f: R \rightarrow R$ be defined as $f( x )=2 x -1$ and $g: R-\{1\} \rightarrow R$ be defined as $g(x)=\frac{x-\frac{1}{2}}{x-1}$ Then the composition function $f(g(x))$ is

R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x - y | ≤ 1. Then, R is:

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi), \operatorname{are}$

If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$

Let $f(x) = (x + |x|) |x|.$ Then, for all $x:$

$\sin {\rm{ }}\left[ {3\,{{\sin }^{ - 1}}\left( {\frac{1}{5}} \right)} \right] = $

Let $f(x) = x^3 -6x^2 +15x + 3$. Then :

Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 00$, and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1^{+}} g(x)=p$, then

Let $A$ is a symmetric and $ \,B$ is a skew symmetric matrix, such that $A - B = \left[ {\begin{array}{*{20}{c}}
1&2 \\
3&4
\end{array}} \right]$, then $|A|$ is