`IfI_(m , n)=intcos^m xsinn xdx ,t h e n7I_(4,3)-4I_(3,2)i se q u a lto` constant (b) `-cos^2x+C` `-cos^4xcos3x+C` (d) `cos7x-cos4x+C`

Options

A. constant
B. `-cos^(2)x+C`
C. `-cos^(4)x cos 3x+C`
D. `cos 7x-cos 4x+C`

Answer

C

Explanation

`I_(4,3)=int cos^(4)x sin3x dx`
Integrating by parts, we have
`I_(4,3)= -(cos3xcos^(4)x)/(3)-(4)/(3)intcos^(3)xsinx cos3x dx`
But `sinx cos3x= -sin2x+sin3x cosx.` So,
`I_(4,3)= -(cosx cos^(4)x)/(3)+(4)/(3)intcos^(3)x sin 2x dx-(4)/(3)intcos^(4)x sin3xdx+C`
`= -(cos3xcos^(4)x)/(3)+(4)/(3)I_(3,2)-(4)/(3)I_(4,3)+C`
Therefore, `(7)/(3)I_(4,3)-(4)/(3)I_(3,2)= -(cos3x cos^(3)x)/(3)+C`
or `7I_(4,3)-4I_(3,2)= -cos 3x cos^(4)x+C`