Evaluate the integral?

Question

sinx/cos^2x dx from 0 to pi/3

This is how it was written in the quiz and I can't get it to go into the calculator written this way...

Answer 1

To put the expression in the calculator you will probably need to input

sin(x)/(cos(x))^2.

It should be something like that for a TI calculator, I don't know much about HP's they use reverse polish notation. Mathematica entry would be

Integrate[Sin[x]/Cos[x]^2, {x,0,Pi/3}].

Writing sin^k(x) or cos^k(x) is just notational convenience, it means the same thing as (sin(x))^k or (cos(x))^k respectively, as long as k ≠ -1.

Note however that the sin^-1(x) and cos^-1(x) have very different meanings, and they are NOT the same as 1/sin(x) = csc(x) or 1/cos(x) = sec(x), they are in fact arcsin(x) and arccos(x).

Anyway, here is how you do the integral if you want to do it by hand.

Let u = cos x.
Observe that du/dx = -sin x.
So you have - du = sin x dx.

Our indefinite integral now becomes
∫ -1/u^2 du = - ∫ u^(-2) du = 1/u + C.

Recall that u = cos x, so our integral becomes 1/(cos x) = sec x.

We substitute in the appropriate values, so our definite integral from 0 to pi/3 should be sec(pi/3) - sec(0).

sec(pi/3) = 1/cos(pi/3) = 1/(1/2) = 2.
sec(0) = 1/cos(0) = 1/1 = 1.

So the final answer should be 2 - 1 = 1.

Answer 2

sinx/cos^2x dx from 0 to pi/3

First rewrite sinx/cos^2x as [sin(x)/cos(x)]*[1/cos(x)]

This simplifies to tan(x)sec(x)

The integral of tan(x)sec(x) dx = sec(x) +C

Since this in a definite integral that is to be evaluated from 0 to pi/3 this can be done by evaluating

sec(pi/3)-sec(0) = 2-1=1

If you want to put the expression into the calculator it needs to be written as sin(x)/(cos(x))^2

Answer 3

Why are you using your calculator to do an integral? Integrals are meant to be done by hand.

∫ sin(x) / cos^2(x) dx

Let u = cos(x)
du = -sin(x) dx
dx = -du / sin(x)

= ∫ -1 / u^2 du
= - ∫ u^(-2) du
= -u^-1 / -1
= 1 / u
= 1 / cos(x)

Now evaluate it at the limits:
[1 / cos(pi/3)] - [1 / cos(0)]
= [1 / (1/2)] - [1/1]
= 2 - 1
= 1

Answer 4

I = ∫ sin x / cos ² x dx ----lims 0 to π/3
Let u = cos x
du = - sin x dx
When x = 0 , u = 1
When x = π / 3, u = 1/2
I = - ∫ du / u ² du----lims 1 to 1/2
I = ∫ du / u ² du----- lims 1/2 to 1
I = ∫ u^(-2) du------- lims 1/2 to 1
I = u^(-1) / (-1)------ lims 1/2 to 1
I = - (1 / u)----------- lims 1/2 to 1
I = - (1 - 2)
I = 1

Answer 5

♦ = -d(cosx) / (cosx)^2, hence I=1/cosx = 2-1= 1; sorry!