The integral `∫(dx/(5 + 4cosx))` is equal to

Answer

`a to p,q; b to r,s; c to p; d to p,q`

Explanation

a. `" Let " I=int(2^(x))/(sqrt(1-4^(x)))dx=(1)/(log2)int(1)/(sqrt(1^(2)-t^(2)))dt`
`"Putting "2^(x)=t, 2^(x)log2dx=dt, " we get "`
`I=(1)/(log2)sin^(-1)((t)/(1))+C=(1)/(log 2)sin^(-1)(2^(x))+C`
` :. k=(1)/(log2)`
b. ` int(dx)/((sqrt(x))^(2)+(sqrt(x))^(7))=int(dx)/((sqrt(x))^(7)(1+(1)/((sqrt(x))^(5))))`
`"Put " (1)/((sqrt(x))^(5))=y, (dy)/(dx)=-(5)/(2(sqrt(x))^(7))`
` :. I=int(-2dy)/(5(1+y))=-(2)/(5)In|1+y|+C=(2)/(5)In((1)/(1+(1)/((sqrt(x))^(5))))`
` :. a=(2)/(5), k=(5)/(2)`
c. Add and subtract ` 2x^(2)` in the numerator. Then `k=1` and ` m=1 `
d. `I=int(dx)/(5+4cosx)`
`=int(dx)/(5("sin"^(2)(x)/(2)+"cos"^(2)(x)/(2))+4("cos"^(2)(x)/(2)-"sin"^(2)(x)/(2)))`
`=int(dx)/(9"cos"^(2)(x)/(2)+"sin"^(2)(x)/(2))=int("sec"^(2)(x)/(2))/(9+"tan"^(2)(x)/(2))dx`
`"Let " t="tan"(x)/(2) " or " 2dt="sec"^(2)(x)/(2)dx`
` :. I=int(2dt)/(9+t^(2))=(2)/(3)"tan"^(-1)((t)/(3))+C`
`=(2)/(3)"tan"^(-1)(("tan"((x)/(2)))/(3))+C`
` :. k=(2)/(3), m=(1)/(3)`