Math, asked by vishalaggarwalsaini, 11 months ago

limit 0 to π/2 ∫tan^8x + tan^10x dx

Answers

Answered by abhi178
1

question is -> \int\limits^{\pi/2}_0{(tan^8x+tan^{10}x)}\,dx

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, \int\limits^{\infty}_0{z^8}\,dz

= \left[\frac{z^9}{9}\right]^{\infty}_0

= ∞ ( it means, limit doesn't converge)

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