Evaluate the definite integral.?

Question

Show your work. thanks.
The integral with lower limit 0 and upper limit pi/3 of sinx / (cosx)^2 dx.

1 is the answer to this question.

Answer 1

Integral (0 to pi/3 , sin(x)/[cos(x)]^2) dx

Note: We have "sin(x)dx" up there, so now our goal is to use substitution to obtain sin(x)dx. We do this by doing the following:

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

Now, we change our boundaries of integration, since we let u = cos(x).
When x = 0, u = 1.
When x = pi/3, u = cos(pi/3) = 1/2.

Our new integral then becomes

Integral (1 to 1/2, 1/(u^2) * (-du))

Pull out all constants (in this case, the negative sign is the constant.

(-1) Integral (1 to 1/2, 1/u^2) du

We're going backwards in the integration, i.e. from 1 to 1/2. We can flip the terms so long as we compensate with a (-1) on the outside of the integral.

(-1) (-1) Integral (1/2 to 1, 1/u^2) du

The two negatives will cancel out.

Integral (1/2 to 1, 1/u^2) du
Integral (1/2 to 1, u^(-2)) du

Then, integration is trivial.

(u^(-1))/(-1) evaluated from 1/2 to 1. Changing the answer around,

-1/u evaluated from 1/2 to 1 is equal to

(-1/1 - (-1/(1/2)) = -1 + 2 = 1.

Answer 2

The integral with lower limit 0 and upper limit pi/3 of sinx / (cosx)^2 dx.
u=cosx
du = -sinx dx

integral sinx /(cosx)^2 dx = integral -du/u^2
=integral -u^-2 du = -u^{-2+1}/-1 = u^{-1} = 1/cosx

with the limits:
1/cos(pi/3) - 1/cos(0) = 1/(1/2) -1 = 2-1 =1 .....

Answer 3

For this problem, you need to use u-substitution:

Let u = cosx, thus du = -sinx*dx

Using these substitutions, the integrand becomes:

-1/u^2*du.

Remember, when doing u-substitution, you also need to change the limits of integration, by subsituting into u = cos(x)

Lower limit: u = cos(0) = 1
Upper limit: u= cos(pi/3) = 1/2

Thus, the intergral becomes:

int(1->1/2): -1/u^2*du. Multiply by -1 to reverse the limits for increasing order.

int(1/2->1): 1/u^2*du

Use the power rule to integrate. Thus you get:

-u^(-1) = -1/u (evaluated from 1/2 to 1)

Evaluating, you get:

-1/(1) - 1/(1/2) -----> -1+2 = 1

-----------

Hope this helps

Answer 4

The integral of sin x / (cos x)^2 is 1/cos x.
You can do that with a substitution; I just know that 1/(cos x)^2 must come from 1/(cos x) somehow, and differentiate that to see if I need to multiply by some constant, which I don't.

Thus, substitute in the limits: 1/cos(pi/3) - 1/cos(0) = 1/0.5 - 1 = 2 - 1 = 1.