The value of the integral `int(cos^3x+cos^5x)/(sin^2x+sin^4x)dxi s` `
` (a).`sinx-6tan^(-1)(sinx)+C` (b). `sinx-2(sinx)^(-1)+C` (c). `sinx-2(sinx)^(-1)-6tan^(-1)(sinx)+C` (d).`sinx-2(sinx)^(-1)+5tan^(-1)(sinx)+C`

Answer

C

Explanation

Let `I=int(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)dx`
`=int((cos^(2)+xos^(4)x)*cos x dx)/((sin^(2)x+sin^(4)x))`
Put `sin x = t rArr cos x dx = dt`
`therefore" "I=int([(1-t^(2))+(1-t^(2))^(2)])/(t^(2)+t^(4))dt`
`rArr" "I=int(1-t^(2)+1-2t^(2)+t^(4))/(t^(2)+t^(4))dt`
`rArr" "I=int(2-3t^(2)+t^(4))/(t^(2)(t^(2)+1))dt" "....(i)`
Using partial fraction for `(y^(2)-3y+2)/(y(y+1))=1+(A)/(y)+(B)/(y=1)" "["where", y = t^(2)]`
`rArr" "A=2, B=-6`
`therefore" "(y^(2)-3y+2)/(y(y+1))=1+(2)/(y)-(6)/(y+1)`
Now, Eq. (i) reduces to, `I=int(1+(2)/(t^(2))-(6)/(1+t^(2)))dt`
`=t-(2)/(t)-6 tan^(-1)(t)+c`
`=sin x -(2)/(sin x)-6 tan^(-1)(sin x)+c`