Question

∫tan^8x + tan^10x dx​

Answer 1

Step-by-step explanation:

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

[\frac{z^9}{9}]^{\infty}_0

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

Answer 2

Step-by-step explanation:

Step-by-step explanation:

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

[\frac{z^9}{9}]^{\infty}_0

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