`"I f"int(dx)/(cos^3xsqrt(sin2x))=a(tan^2x+b)sqrt(tanx)+c`

Options

A. `a=(sqrt(2))/(5), b=(1)/(sqrt(5))`
B. `a=(sqrt(2))/(5),b=5`
C. `a=(sqrt(2))/(5),b=-(1)/(sqrt(5))`
D. `a=(sqrt(2))/(5), b=sqrt(5)`

Answer

B

Explanation

We have , `int(dx)/(cos^(3)xsqrt(sin2x))`
`=a(tan^(2)x+b)sqrt(tanx)+c`
Let `I=(1)/(sqrt(2))int(dx)/(cos^(3)xsqrt(sinxcosx))`
`=(1)/(sqrt(2))intsec^(4)x(dx)/(sqrt(tanx))`
`=(1)/(sqrt(2))int((1+tan^(2)x)sec^(2)x)/(sqrt(tanx))dx`
Put tan x = t
`implies" "sec^(2)x" dx"=dt`
`:." "l=(1)/(sqrt(2))int((1+t^(2)))/(sqrt(t))dt`
`=(1)/(sqrt(2))int(t^(-1//2)+t^(3//2))dt`
`=(1)/(sqrt(2))(2t^(1//2)+(2)/(5)t^(5//2))+c`
`=(1)/(sqrt(2))[2sqrt(tanx)+(2)/(5)(tanx)^(5//2)]+c`
`=(sqrt(2)+(sqrt(2))/(5)tan^(2)x)sqrt(tanx)+c`
`implies" "a(tan^(2)x+b)sqrt(tanx)+c`
`impliesa=(sqrt(2))/(5),ab=sqrt(2)`
`impliesb=5`