limit 0 to π/2 ∫tan^8x + tan^10x dx
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first of all, resolve the (tan^8x + tan^10x) in simpler form.
i.e., tan^8x + tan^10x = tan^8x(1 + tan²x)
we know, sec²θ - tan²θ = 1
so, 1 + tan²x = sec²x
hence, tan^8x + tan^10x = sec²x . tan^8x
let tanx = z
differentiating both sides,
sec²x dx = dz
upper limit : z = tan(π/2) = ∞
lower limit : z = tan(0) = 0
so,
=
= ∞ ( it means, limit doesn't converge)
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