- e sin x-1-x1-x cos x-1-x d x isA- cos x - esin x-1-x-cB- e sin x-1-x- x - cC- 1-x- esin -1-x-cD- x - e sin x -1-x- c

The correct option is D x.e^(sin x 1/x)+c∫ e^(sin x 1/x) (1 + x cos x + 1/x) dx = ∫e^(sin x 1/x) dx+ ∫x . e^(sin x 1/x) (cos x + 1/x^2)dx = ∫1 · e^(sin ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

∫ e sin x-1/x...

Question

$\int e^{(sin x - \frac{1}{x})} (1 + x cos x + \frac{1}{x}) d x$ is

cos x . e^{(sin x - \frac{1}{x})} + c

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\frac{1}{x} . e^{(sin - \frac{1}{x})} + c

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x . e^{(sin x - \frac{1}{x})} + c

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e^{(sin x - \frac{1}{x})} + x + c

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Solution

The correct option is C $x . e^{(sin x - \frac{1}{x})} + c$
$\int e^{(sin x - \frac{1}{x})} (1 + x cos x + \frac{1}{x}) d x$
$= \int e^{(sin x - \frac{1}{x})} d x + \int x . e^{(sin x - \frac{1}{x})} (cos x + \frac{1}{x^{2}}) d x$
$= \int 1 \cdot e^{(sin x - \frac{1}{x})} d x + \int x . e^{(sin x - \frac{1}{x})} (cos x + \frac{1}{x^{2}}) d x$
$= x . e^{(sin x - \frac{1}{x})} - \int x . e^{(sin x - \frac{1}{x})} (cos x + \frac{1}{x^{2}}) d x + \int x . e^{(sin x - \frac{1}{x})} (cos x + \frac{1}{x^{2}}) d x + c$
$= x . e^{(sin x - \frac{1}{x})} + c$

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∫e(sinx−1x)(1+x cosx+1x)dx is

The correct option is C x.e(sin x−1x)+c∫e(sinx−1x)(1+x cosx+1x)dx =∫e(sinx−1x)dx+∫x.e(sinx−1x)(cosx+1x2)dx =∫1⋅e(sinx−1x)dx+∫x.e(sinx−1x)(cosx+1x2)dx =x.e(sinx−1x)−∫x.e(sinx−1x)(cosx+1x2)dx+∫x.e(sinx−1x)(cosx+1x2)dx+c =x.e(sinx−1x)+c

$\int e^{(sin x - \frac{1}{x})} (1 + x cos x + \frac{1}{x}) d x$ is