Question #5adb3

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Answer 1 · 2016-09-20T20:47:12

Alternative I :-

$-{2cosx+ln|cscx-cotx|+C.$

Alternative II :-

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

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

Alternative I :-

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

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

$=2intsinxdx-intcscxdx$

$=2(-cosx)-ln|cscx-cotx|$

$=-{2cosx+ln|cscx-cotx|+C.$

Alternative II :-

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

We use the Identity $: sin^2x=(1-cos2x)/2$ .

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

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

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

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

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

Enjoy Maths.!