Question #5adb3

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Question #5adb3

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Answer 1 · 2016-09-20T20:47:12

Alternative I :-

$- {2 cos x + ln | csc x - cot x | + C .$ $-{2cosx+ln|cscx-cotx|+C.$

Alternative II :-

$\frac{1}{4} (2 x - sin 2 x) - ln | csc x - cot x | - cos x + C$ $1/4(2x-sin2x)-ln|cscx-cotx|-cosx+C$

Explanation:

Though the Question is not clear, but we solve it by taking $2$ $2$ possibilities :

Alternative I :-

$\int \frac{{sin}^{2} x - {cos}^{2} x}{sin x} d x = \int {\frac{{sin}^{2} x - (1 - {sin}^{2} x)}{sin x} d x$ $int(sin^2x-cos^2x)/sinx dx=int{(sin^2x-(1-sin^2x)}/sinxdx$

$= \int \frac{2 {sin}^{2} x - 1}{sin x} d x = \int {\frac{2 {sin}^{2} x}{sin x} - \frac{1}{sin x}} d x$ $=int(2sin^2x-1)/sinx dx=int{(2sin^2x)/sinx-1/sinx}dx$

$= 2 \int sin x d x - \int csc x d x$ $=2intsinxdx-intcscxdx$

$= 2 (- cos x) - ln | csc x - cot x |$ $=2(-cosx)-ln|cscx-cotx|$

$= - {2 cos x + ln | csc x - cot x | + C .$ $=-{2cosx+ln|cscx-cotx|+C.$

Alternative II :-

$\int {{sin}^{2} x - \frac{{cos}^{2} x}{sin x}} d x$ $int{sin^2x-cos^2x/sinx}dx$

We use the Identity $: {sin}^{2} x = \frac{1 - cos 2 x}{2}$ $: sin^2x=(1-cos2x)/2$ .

$:. int{sin^2x-cos^2x/sinx}dx$

$=int(1-cos2x)/2dx-int(1-sin^2x)/sinxdx$

$1/2(x-sin(2x)/2)-int(1/sinx-sinx)dx$

$=1/4(2x-sin2x)-intcscxdx+intsinxdx$

$=1/4(2x-sin2x)-ln|cscx-cotx|-cosx+C$

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Question #5adb3

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