Help with calculus problem!

i've been working on this problem for ages and i can't figure it out! can someone give me some guidance please?

it's the integral of sec^10(x)tan(x)dx

thank you!

Possible intermediate steps:
integral sec^10(x) tan(x) dx
For the integrand tan(x) sec^10(x), use the trigonometric identity sec^2(x) = tan^2(x)+1:
= integral tan(x) (tan^2(x)+1)^4 sec^2(x) dx
For the integrand tan(x) (tan^2(x)+1)^4 sec^2(x), substitute u = tan(x) and du = sec^2(x) dx:
= integral u (u^2+1)^4 du
For the integrand u (u^2+1)^4, substitute s = u^2+1 and ds = 2 u du:
= 1/2 integral s^4 ds
The integral of s^4 is s^5/5:
= s^5/10+constant
Substitute back for s = u^2+1:
= 1/10 (u^2+1)^5+constant
Substitute back for u = tan(x):
= (sec^10(x))/10+constant

Its simple...
First simplify in terms of sin and cos..
So,integral of(sin x/cos^10)dx
now,put cos x=t,
so,-sin x=dt,
so,your problem becomes,, integral of(-t^10)dt
by solving,you will get,10t^-9
now,put t=cos x and your answer will be 10 sec ^9 x....

Write the integral as
∫ sec^9(x) sec(x) tan(x) dx.
Now recall that the derivative of
sec(x) is sec(x)*tan(x) dx.
So let u = sec(x) du = sec(x)*tan(x) dx to get
∫ u^9 du = u^10/10 + C.
Finally back substitute to get the final answer:
sec^10(x)/10 + C.

Rewrite that as
∫sec^9(x)*sec(x)tan(x)dx
Set u = sec(x)
du = sec(x)tan(x)dx
∫u^9du =
1/10u^10 +C
plug sec(x) back in for u
1/10sec^10(x) + C

∫ sec^10(x) tan x dx

= ∫sec^9(x) sec x tan x dx

let sec x = u

=> sec x tan x dx = du

= ∫ u^9 du

= (1/10)u^10 + C

= (1/10) sec^10(x) + C

Hint: let u = sec(x) → (1/u) du = tan(x) dx