integrate I= cos2x.tan³(x)

 rewrite it as:

∫ (tan³x /sec²x) dx =

split the numerator as:

∫ (tan²x /sec²x) tanx dx =

replace tan²x with (sec²x - 1):

∫ [(sec²x - 1) /sec²x] tanx dx =

divide and multiply the integrand by secx (so as to get secx tanx that is the derivative of secx):

∫ [(sec²x - 1) /sec³x] secx tanx dx =

let secx = u

differentiate both sides:

d(secx) = du →

secx tanx dx = du

substituting, you get:

∫ [(sec²x - 1) /sec³x] secx tanx dx = ∫ [(u² - 1) /u³] du =

distributing, you get:

∫ [(1/u) - (1 /u³)] du =

∫ (1/u) du - ∫ u^(-3) du =

ln |u| - [1/(-3+1)] u^(-3+1) + c =

ln |u| - [1/(-2)] u^(-2) + c =

ln |u| + (1/2)(1/u²) + c 

then, substituting back u = secx, you get:

ln |secx| + (1/2)(1/sec²x) + c

thus, finally:

∫ cos²x tan³x dx = ln |secx| + (1/2) cos²x + c

niladry see the question clearly  i is I=cos2x.tan^3(x)

 oh.. it's easy then

First, bring it in this form:

Now substitute cosx=z in both the terms..

Therefore, we get,

solving further, we get,

Putting z=cosz, we get