Q44 of 865 Page 20

Evaluate the following integrals:

∫ sec4 x tan x dx

Put tanx = t


d(tanx) = dt


sec2xdx = dt


⇒


We can write sec4x = sec2x. sec2x


Now ,the question becomes


⇒


⇒


Tan2x + 1 = sec2x


tanx = t


t2 + 1 = sec2x


⇒


⇒


⇒


But t = tanx


⇒


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