If - xexcos xdx-aexb1-xsin x-cxcos x-d- then

The correct option is B a=1/2, b= 1, c=1 Given nbsp; ∫ xe^xcos xdx=ae^x(b(1 x)sin x+cxcos x)+d nbsp; nbsp; nbsp; ....(1)Consider, I=∫ ...

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Standard XII

Mathematics

Principal Solution of Trigonometric Equation

If ∫ xexcos...

Question

If $\int x e^{x} cos x d x = a e^{x} (b (1 - x) sin x + c x cos x) + d,$ then

a = - 1, b = 1, c = - 1

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a = \frac{1}{2}, b = - 1, c = 1

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a = 1, b = - 1, c = 1

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a = \frac{1}{2}, b = - 1, c = - 1

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Solution

The correct option is B $a = \frac{1}{2}, b = - 1, c = 1$
Given $\int x e^{x} cos x d x = a e^{x} (b (1 - x) sin x + c x cos x) + d$ ....(1)
Consider, $I = \int x e^{x} cos x d x$
$= x e^{x} sin x - \int (x e^{x} + e^{x}) sin x d x$
$= x e^{x} sin x - \int x e^{x} sin x d x - \int e^{x} sin x d x$
$= x e^{x} sin x - x e^{x} (- cos x) - \int (x e^{x} + e^{x}) cos x d x - \int e^{x} sin x d x$
$= x e^{x} sin x + x e^{x} (cos x) - \int x e^{x} cos x d x - \int e^{x} (cos x + sin x) d x$
or $2 I = x e^{x} (sin x + cos x) - e^{x} sin x + d$ ( $∵ \int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C$ )
$= e^{x} ((x - 1) sin x + x cos x) + d$
or $I = \frac{1}{2} e^{x} ((x - 1) sin x + x cos x) + d$
So,by comparing with (1), we get
$a = \frac{1}{2}, b = - 1, c = 1$

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If ∫xexcosxdx=aex(b(1−x)sinx+cxcosx)+d, then

If $\int x e^{x} cos x d x = a e^{x} (b (1 - x) sin x + c x cos x) + d,$ then