# What is primitive of function (cosx)^3/sin x?

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### 1 Answer

You need to use the evaluate the primitive of the given function, hence, you need to evaluate the indefinite integral of the given function, such that:

`int (cosx)^3/sin x dx = int (cos^2 x)/sin x *cos x dx`

You need to use the Pyrhagorean trigonometric identity, such that:

`cos^2 x = 1 - sin^2 x`

`int (cos^2 x)/sin x *cos x dx = int (1 - sin^2 x)/sin x *cos x dx`

You may solve the indefinite integral using the substitution process, such that:

`sin x = y => cos x dx = dy`

Replacing the variable yields:

`int (1 - sin^2 x)/sin x *cos x dx = int (1 - y^2)/y *dy`

You need to split the integral using the property of linearity of integral, such that:

`int (1 - y^2)/y *dy = int 1/y dy - int y dy`

`int (1 - y^2)/y *dy = ln|y| - y^2/2 + c`

Replacing back `sin x` for `y` yields:

`int (cosx)^3/sin x dx = ln|sin x| - (sin^2 x)/2 + c`

Hence, evaluating the requested primitive yields, under the given conditions, `int (cosx)^3/sin x dx = ln|sin x| - (sin^2 x)/2 + c.`

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