_ ^ 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 7.1 inderfinite integral→7.1 inderfinite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\alpha \in(0,1)$ and $\beta=\log _{ c }(1-\alpha)$. Let $P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots . .+\frac{x^n}{n}, x \in(0,1)$ Then the integral $\int \limits_0^\alpha \frac{ t ^{50}}{1- t } dt$ is equal to

Let $f(x) = x^2, x \in R$. For any $A \subseteq R$, define $g(A) = \{x \in R : f(x) \in A\}$. If $S = [0, 4]$, then which one of the following statements is not true ?

Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

Evaluate: $\int \frac{\cot x}{\sqrt[3]{\sin x}} d x$

If $y = {\sin ^{ - 1}}\left( {{{19} \over {20}}x} \right) + {\cos ^{ - 1}}\left( {{{19} \over {20}}x} \right)$, then ${{dy} \over {dx}} = $

If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of

$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$

Find the principal values of: $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$

$\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$ is equal to:

$\text{x}+\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{x}+\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{x}-\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$

Which of the following is correct?

Determinant is a square matrix.
Determinant is a number associated with a matrix.
Determinant is a number associated with a square matrix.
None of these.

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) d x=\left(\int_0^1 f(x) d x\right)^2$. Then, the range of $f$