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`y' = sqrt(tanx)sec^4x` Solve the differential equation.

Given to solve


`y' = sqrt(tan x) *(sec^4 x)`


=> `y = int sqrt(tan x) *(sec^4 x) dx`


so ,


we have to solve


`int sqrt(tan x) *(sec^4 x) dx`


= `int sqrt(tan x) *(sec^2 x) *(sec^2 x) dx`


= `int (1+tan^2 (x))*(sec^2 x)*(sqrt(tan x)) dx`


let ,


`u= tan x`


so, `du = sec^2 (x) dx`


so ,


`int (1+tan^2 (x))*(sec^2 x)*(sqrt(tan x)) dx`


=` int (1+u^2)(u^(1/2)) du`


= `int (u^(1/2)+u^(5/2)) du`


=` (u^((1/2) +1))/((1/2)+1) +(u^((5/2) +1))/((5/2)+1) `


= `(u^((3/2) ))/((3/2)) +(u^((7/2) ))/((7/2))`


= `2/3u^(3/2) +2/7u^(7/2)`


= `2/3(tan x)^(3/2 ) +2/7(tan x)^(7/2)`

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