The correct option is B #10; x^2/4 xsin2x/4 1/8cos2x +c #10; ∫ x sin^2x dx. = x ∫ sin^2x. dx ∫ ( ∫ sin^2x.dx )dx cos 2x= 1 2 sin ...

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Domain and Range of Basic Inverse Trigonometric Functions

∫ xsin2xdx

Question

$\int x s i n^{2} x d x$

\frac{x^{2}}{4} - \frac{cos 2 x}{3} + \frac{1}{8} s i n 2 x + c

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\frac{x^{2}}{4} - \frac{x s i n 2 x}{4} - \frac{1}{8} c o s 2 x + c

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\frac{x^{2}}{4} + \frac{x s i n 2 x}{4} + \frac{1}{8} c o s 2 x + c

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\frac{- x^{2}}{4} - \frac{c o s 2 x}{3} + \frac{1}{8} c o s 2 x + c

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Solution

The correct option is B $\frac{x^{2}}{4} - \frac{x s i n 2 x}{4} - \frac{1}{8} c o s 2 x + c$
$\int x s i n^{2} x d x .$
$= x \int s i n^{2} x . d x$ $- \int (\int s i n^{2} x . d x) d x$
$c o s 2 x = 1 - 2 s i n^{2} x$
$2 s i n^{2} x = \frac{1 - c o s 2 x}{2}$
$x \int (\frac{1 - c o s 2 x}{2}) d x - \int (\int (\frac{1 - c o s 2 x}{2}) . d x) d x .$
$= \frac{x}{2} [x -^{1} /_{2} s i n 2 x] - \int [\frac{x}{2} -^{1} /_{2} s i n 2 x] d x$
$= \frac{x^{2}}{2} - \frac{x}{4} s i n 2 x - \int [\frac{x^{2}}{4} + \frac{1}{4} c o s 2 x]$
$= \frac{x^{2}}{4} - \frac{x s i n 2 x}{4} - \frac{1}{8} c o s 2 x + c$

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