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Int (sinx)^4/(cosx)6

Int (sinx)^4/(cosx)6

Grade:12

2 Answers

Sharandeep singh
15 Points
9 years ago

this is  (sinx/cosx)^4 / cosx ^2

= (tanx)^4 * (secx)^2

put tanx = t

diff. sec^2x dx = dt

we get  (t)^4 dt    ( sec^2x dx = dt  and tanx = t)

 = t^5/5

= tanx^5 / 5

renu akunuri
37 Points
8 years ago

int tan x4 .sec x2  dx

=int tan x4 (1+tan x2) dx

=int tan x4 dx +int tan x6 dx                  it is the formula

=(tan x3) /3 -tan x +(tan x5 )/5 -(tan x3  )/3 +tan x

=(tan x5) /5

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