∫xI cos xI.dx Using by parts ∫ x cos x.dx = x ∫cos x.dx ∫ nbsp; [ ddx(x)∫cos x.dx ].dx = x [sin x] ∫ 1.sin x.dx = x sin x ( cos x)+C = x sin x + cos x+ C ...

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Standard IX

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Machiavelli

Find: ∫ x cos...

Question

Find: $\int x cos x . d x$

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Solution

$\int x    I cos x      I . d x$
Using by parts
$\int x cos x . d x$
$= x \int cos x . d x - \int [\frac{d}{d x} (x) \int cos x . d x] . d x$
$= x [sin x] - \int 1. sin x . d x$
$= x sin x - (- cos x) + C$
$= x sin x + cos x + C$
Where $C$ is constant of integration.

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Machiavelli

Standard IX History

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Find: ∫x cos x.dx

∫xI cosxI.dx Using by parts ∫x cos x.dx =x∫cos x.dx−∫[ddx(x)∫cos x.dx].dx =x[sin x]−∫1.sin x.dx =x sin x−(−cos x)+C =x sin x+cos x+C Where C is constant of integration.

Find: $\int x cos x . d x$

$\int x    I cos x      I . d x$
Using by parts
$\int x cos x . d x$
$= x \int cos x . d x - \int [\frac{d}{d x} (x) \int cos x . d x] . d x$
$= x [sin x] - \int 1. sin x . d x$
$= x sin x - (- cos x) + C$
$= x sin x + cos x + C$
Where $C$ is constant of integration.