Q. 445.0( 1 Vote )

# 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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