# Find the derivative of the function.Find the derivative of the function. y= integral (cosx) to (sinx) ln(4+3v)dv y'(x)=__________________?

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### 1 Answer

You need to use integration by parts to evaluate the integral such that:

`int fg' = fg - int f'g`

`f= ln(4+3v) => f' = 3/(4+3v) dv`

`g' = dv => g = v`

`int_cos x^sin xln(4+3v) dv = vln(4+3v)|_cos x^sin x - int_cos x^sin x (3v)/(4+3v) dv`

You need to evaluate the integral `int_cos x^sin x (3v)/(4+3v) dv` , hence, you need to add and subtract 4 to numerator such that:

`int_cos x^sin x (3v+4-4)/(4+3v) dv = int_cos x^sin x (3v+4)/(4+3v) dv - int_cos x^sin x (4)/(4+3v) dv`

`int_cos x^sin x (3v+4-4)/(4+3v) dv = int_cos x^sin x dv - int_cos x^sin x (4)/(4+3v) dv`

You should use the following substitution to solve `int_cos x^sin x (4)/(4+3v) dv` such that:

`4+3v = u => 4dv = du => dv = (du)/3`

`int (4)/(4+3v) dv = (4/3)int (du)/u = (4/3) ln|u| + c`

Substituting back `4+3v` for u yields:

`int_cos x^sin x (4)/(4+3v) dv = (4/3) ln|4+3v||_cos x^sin x`

`int_cos x^sin x (3v+4-4)/(4+3v) dv = v|_cos x^sin x - (4/3) ln|4+3v||_cos x^sin x`

`int_cos x^sin x ln(4+3v) dv = vln(4+3v)|_cos x^sin x-v|_cos x^sin x+ (4/3) ln|4+3v||_cos x^sin x`

`int_cos x^sin x ln(4+3v) dv = sin x*ln(4+3sin x) - sin x + (4/3) ln|4+3sin x| - cos x*ln(4+3cos x) + cos x - (4/3) ln|4+3cos x|`

Hence, evaluating y(x) yields:

`y(x) = sin x*ln(4+3sin x) - sin x + (4/3) ln|4+3sin x| - cos x*ln(4+3cos x) + cos x - (4/3) ln|4+3cos x|`

You need to use chain rule and product rule to evaluate y'(x) such that:

`y'(x) = cos x*ln(4+3sin x) + sin x*((3cos x)/(4+3sin x)) - cos x + ((4cos x)/(4+3sin x)) + sin x*ln(4+3cos x) +cos x*((3sin x)/(4+3cos x))- sin x + ((4sin x)/(4+3cos x))`

**Hence, evaluating the derivative of the function y(x) yields `y'(x) = cos x*ln(4+3sin x) + sin x*((3cos x)/(4+3sin x)) - cos x + ((4cos x)/(4+3sin x)) + sin x*ln(4+3cos x) + cos x*((3sin x)/(4+3cos x))- sin x + ((4sin x)/(4+3cos x)).` **