evaluate the integral?

Question

[sin(x) + sin(x)tan^2(x)] / sec^2(x) dx

on [0,pi/3]

so far i've gotten
=sin(x)(1 + tan^2(x))/sec^(x)
=sin(x)(sec^2)/sec^2(x)

and im stuck from here.

Answer 1

[sin(x) + sin(x)tan²(x)] / sec²(x)
= sin(x)[1 + tan²(x)] / sec²(x)
= sin(x)sec²(x)/sec²(x)
= sin(x)

∫ [sin(x) + sin(x)tan²(x)] / sec²(x) dx
= ∫ sin(x) dx
= -cos(x)

Apply the limits to get:
-cos(π/3) - (-cos(0))
= -1/2 - (-1)
= -1/2 + 1
= 1/2

Answer 2

sin x + sin tan ² x
sin x (1 + tan ² x)
sin x sec ² x

I = ∫ sin x dx
I = - cos x between 0 and π/3
I = - (1/2 - 1)
I = 1/2

Answer 3

u got most or it done (assuming u mean (secx)^2 i guessing u do since u used 1+(tanx)^2 = (secx)^2

the two (secx)^2 cancle and ur left with:
integral of sinx from 0 to pi/3
= -cosx => -cos(pi/3) + cos 0 = -1/2 + 1 = 1/2