If `int(dx)/(root3(sin^(11)xcosx))=-[(3)/(8)f(x)+(3)/(2)g(x)]+C`, then

Options

A. `f(x)=tan^(-8//3)x, g(x)=tan^(-2//3)x`
B. `f(x)=tan^(8//3)x, g(x)=tan^(-2//3)x`
C. `f(x)=tan^(-8//3)x, g(x)=tan^(2//3)x`
D. `f(x)tan^(4//3)x, g(x)=tan^(-4//3)x`

Answer

A

Explanation

Sum of powers `=(-11)/(3)-(1)/(3)=-4" [negative and even]"`
`"Put "tanx=t`
`rArr" "sec^(2)xdx=dt`
`thereforeint(dx)/(root3(sin^(11)x.cosx))=int(dx)/(root3((sin^(11)x)/(cos^(11)x)).cos^(4)x)=int(sec^(4)xdx)/(root3(tan^(11)x))`
`=int(1+t^(2))/(t^(11//3))dt=int(t^(-11//3)+t^(-5//3)dt)`
`=-((3)/(8)tan^(-8//3)x+(3)/(2)tan^(-2//3)x)+C`
`rArr" "f(x)=tan^(-8//3)x and g(x)=tan^(-2//3)x`