Evalutating the integral?

Question

evalute the integral

intergrate from0 to pie/3 (sinx+ sinx tan2x)/ sec2x

trig is not my strong point can someone help he with this

Answer 1

Buri is corrct if it is (tanx)^2 and (secx)^2

If it is tan2x and sec2x, it will be different.

In that case

(sinx+ sinx tan2x)/ sec2x

(sinx+ sinx tan2x)cos2x

=sinx cos2x + sinxsin2x

=(1/2){ sin3x - sinx} +(1/2) {cosx - cos3x} and then integrate

Integral = (1/2){- (c0s3x)/3 +cosx} + (1/2){ sinx -(sin3x)/3}

and evaluate between the given limits.

Answer 2

∫ (sin(x) + sin(x)tan²(x))/sec²(x) dx

factor out sin(x)

=∫ [sin(x) ][1 + tan² (x)]/sec²(x) dx

recall that sec²(x) = 1 + tan²(x)

thus,

=∫ [sin(x) ][sec²(x)]/sec²(x) dx

=∫ sin(x) dx = -cos(x) with limits from 0 to pi/3

=[-cos(pi/3)] - [-cos(0)]

=[-(1/2)+ 1]

=1/2

=)

Answer 3

Simplify the function your integrating:

[sin(x) + sin(x)tan²(x)]/sec²(x)

= [sin(x)(tan²(x) + 1)]/sec²(x); after factoring out the sin(x)

= [sin(x)sec²(x)]/sec²(x); because of the identity tan²(x) + 1 = sec²(x)

= sin(x)

Therefore,

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

Now evaluating at the bounds:

∫ [0, π/3] sin(x) dx = [0, π/3] -cos(x) = -cos(π/3) - (-cos(0)) = -(1/2) + 1 = 1/2

Hope this helps!