Help With a Calculus Problem?

If we're going to replace x with 2 tan y (we should use a different variable, otherwise we're saying x = 2 tan x, which limits x to certain values), then we need to evaluate dx as well.

To do that, we use dx/dy = d/dy (2 tan y), and then multiply both sides by dy.

In order to take the derivative of 2 tan y with respect to y, I'll make use of the trig relation tan y = sin y / cos y.

d/dy (2 tan y) = 2 d/dy (sin y / cos y)

Now, because I can only remember the product rule and never the quotient rule,

d/dy (2 tan y) = 2 d/dy [sin y * (cos y)^-1]

d/dy (2 tan y) = 2 {d/dy (sin y) * (cos y)^-1 + sin y d/dy [(cos y)^-1]}

Now, from the chain rule and power rule,

d/dy (2 tan y) = 2 {d/dy (sin y) * (cos y)^-1 + sin y * [-1 * (cos y)^-2 * d/dy (cos y)]}

You should know that the deriviative of sine is cosine, and the derivative of cosine is -sine.

d/dy (2 tan y) = 2 {cos y * (cos y)^-1 + sin y * [-1 * (cos y)^-2 * -sin y]}

d/dy (2 tan y) = 2 {cos y / cos y + sin y * [(cos y)^-2 * sin y]}

d/dy (2 tan y) = 2 {1 + (sin y)^2 / (cos y)^2}

d/dy (2 tan y) = 2 [(cos y)^2 + (sin y)^2] / (cos y)^2

Now, from the Pythagorean theorem, we know sin^2 y + cos^2 y = 1, so

d/dy (2 tan y) = 2 * 1 / (cos y)^2

d/dy (2 tan y) = 2 / (cos y)^2

Now, we go back to

dx/dy = d/dy (2 tan y)

dx/dy = 2 / (cos y)^2

dx = 2 / (cos y)^2 * dy

Now, back to the original integral. Replacing x with 2 tan x, and dx with 2 / (cos y)^2 * dy,

∫1 / (x^2 + 4) dx = ∫1 / [(2 tan y)^2 + 4] * 2 / (cos y)^2 * dy

∫1 / (x^2 + 4) dx = 2 ∫1 / {[(2 tan y)^2 + 4] * (cos y)^2} * dy

∫1 / (x^2 + 4) dx = 2 ∫1 / {[4 (tan y)^2 + 4] * (cos y)^2} * dy

∫1 / (x^2 + 4) dx = 2 / 4 * ∫1 / {[(tan y)^2 + 1] * (cos y)^2} * dy

∫1 / (x^2 + 4) dx = 1/2 * ∫1 / {[(tan y)^2 + 1] * (cos y)^2} * dy

Now, replacing tan y with sin y / cos y,

∫1 / (x^2 + 4) dx = 1/2 * ∫1 / {[(sin y / cos y)^2 + 1] * (cos y)^2} * dy

∫1 / (x^2 + 4) dx = 1/2 * ∫1 / {[(sin y)^2 / (cos y)^2 + 1] * (cos y)^2} * dy

∫1 / (x^2 + 4) dx = 1/2 * ∫1 / [(sin y)^2 + (cos y)^2] * dy

Using the Pythagorean theorem,

∫1 / (x^2 + 4) dx = 1/2 * ∫1 / 1 * dy

∫1 / (x^2 + 4) dx = 1/2 * ∫dy

∫1 / (x^2 + 4) dx = 1/2 * y + C

Don't forget, since this is an indefinite integral, we need to add an unknown constant term. Now, replace y with x. Recall that x = 2 tan y, so y = arctan (x / 2)

∫1 / (x^2 + 4) dx = 1/2 * arctan (x / 2) + C

I hope that helps!

f (x) = x^3 - 12x + a million . . . the 1st spinoff set to 0 famous turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2nd spinoff evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== beneficial fee shows x=2 is a interior sight minimum f ' ' (-2) = 6*(-2) = -12 <== damaging fee shows x=-2 is a interior sight optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is reducing x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2nd spinoff set to 0 famous inflection factors, or the place concavity ameliorations 6x = 0 x = 0 <=== inflection element x = - 2 is a optimal, so could desire to be concave down concavity ameliorations on the inflection element(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

x = 2 * tan(t) would be a better substitution

x = 2 * tan(t)

dx = 2 * sec(t)^2 * dt

dx / (x^2 + 4) =>

(2 * sec(t)^2 * dt) / ((2 * tan(t))^2 + 4) =>

(2 * sec(t)^2 * dt) / (4 * tan(t)^2 + 4) =>

(2 * sec(t)^2 * dt) / (4 * (1 + tan(t)^2)) =>

(2 * sec(t)^2 * dt / (4 * sec(t)^2) =>

(1/2) * dt

Integrate

(1/2) * t + C

x = 2 * tan(t)

x/2 = tan(t)

t = arctan(x/2)

(1/2) * arctan(x/2) + C

∫ 1/(x² + 4) dx

x = 2 tan t

t = arctan (x/2)

dx = 2 sec² t dt

∫ 1/(x² + 4) dx

= ∫ 1/((2 tan t)² + 4) (2 sec² t dt)

= ∫ 1/(4 tan² t + 4) (2 sec² t dt)

= ∫ 1/(4(tan² t + 1)) (2 sec² t dt)

= ∫ 1/(4 sec² t) (2 sec² t dt)

= ∫ (1/2) dt

= (1/2)t + C

= (1/2)(arctan (x/2)) + C

Substitute x = 2*tan(t):

∫2*sec²(t) dt/4*sec²(t) =t/2 + C = arctan(x/2) + C