If - x3-2x2-3x-1 cos 2x dx -sin 2x-4u x -cos 2x-8v x -C then

The correct option is A u ( x )=2x^3 4x^2+3x ∫ ( x^3 2x^2+3x 1 )cos 2x dx #160; =sin 2x/4u ( x )+cos 2x/8v ( x )+C .... (i) Applying ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If ∫ x3-2x2...

Question

If $\int (x^{3} - 2 x^{2} + 3 x - 1) cos 2 x d x$
$= \frac{sin 2 x}{4} u (x) + \frac{cos 2 x}{8} v (x) + C$ then

u (x) = x^{3} - 4 x^{2} + 3 x

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u (x) = 2 x^{3} - 4 x^{2} + 3 x

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u (x) = 3 x^{3} - 4 x + 3

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v (x) = 6 x^{2} - 8 x

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Solution

The correct option is A $u (x) = 2 x^{3} - 4 x^{2} + 3 x$
$\int (x^{3} - 2 x^{2} + 3 x - 1) cos 2 x d x$ $= \frac{sin 2 x}{4} u (x) + \frac{cos 2 x}{8} v (x) + C$ .... $(i)$

Applying the formula given in the comprehension, required integral becomes
$\int (x^{3} - 2 x^{2} + 3 x - 1) cos 2 x d x$
$= (x^{3} - 2 x^{2} + 3 x - 1) \frac{sin 2 x}{2} - (3 x^{2} - 4 x + 3) (- \frac{cos 2 x}{4}) + (6 x - 4) (- \frac{sin 2 x}{8}) - 6 \frac{cos 2 x}{16} + C$

$= \frac{sin 2 x}{4} (2 x^{3} - 4 x^{2} + 6 x - 2) - (3 x - 2) \frac{sin 2 x}{4} + \frac{cos 2 x}{8} (6 x^{2} - 8 x + 6) - 3 \frac{cos 2 x}{8} + C$

$= \frac{sin 2 x}{4} [2 x^{3} - 4 x^{2} + 3 x] + \frac{cos 2 x}{8} [6 x^{2} - 8 x + 3] + C$

Comparing with the given equation $(i)$ , we get
$u (x) = 2 x^{3} - 4 x^{2} + 3 x$

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Similar questions

Q. If

\int (x^{3} - 2 x^{2} + 3 x - 1) c o s 2 x d x = \frac{s i n 2 x}{4} u (x) + \frac{c o s 2 x}{8} v (x) + c

, then

Q. Evaluate the following:

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

Q. If

\int (3 x^{2} + x - 2) {sin}^{2} (3 x + 1) d x

= - \frac{u (x)}{72} sin (6 x + 2) + \frac{v (x)}{72} cos (6 x + 2) + \frac{1}{2} x^{3} + \frac{1}{4} x^{2} - x + C

What must be added to $(x^{3} + 3 x - 8)$ to get $(3 x^{3} + x^{2} + 6)$ ?

Question 24
Factorise
(i) $2 x^{3} - 3 x^{2} - 17 x + 30$

(ii) $x^{3} - 6 x^{2} + 11 x - 6$

(iii) $x^{3} + x^{2} - 4 x - 4$

(iv) $3 x^{3} - x^{2} - 3 x + 1$

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If ∫(x3−2x2+3x−1)cos2xdx=sin2x4u(x)+cos2x8v(x)+C then

If $\int (x^{3} - 2 x^{2} + 3 x - 1) cos 2 x d x$
$= \frac{sin 2 x}{4} u (x) + \frac{cos 2 x}{8} v (x) + C$ then