Integrating Trigonometric Functions with Pythagorean Identities

[geft]

Given

R = \sqrt{d^2 + z^2}

and

\cos \alpha = \frac{d}{R}

find

\int \frac{1}{d^2 + z^2}dz

The answer given is

\frac{1}{d}\tan^{-1}(\frac{z}{d})

I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.

 

[dextercioby]

You're question doesn't make too much sense. I mean if you're asked to find the antiderivative of a function (1/(d^2 + z^2)), what do R and cos alpha have to do with it ?

 

[geft]

Without them, I don't think you'd able to tell which variable is on which side of the triangle, because the answer is in trigonometric form. Also, the question is part of a bigger problem, so please disregard the choice of variables used.

 

[dextercioby]

What triangle ? Why don't you post your problem's text along with your attempts to solve it ?

 

[Mark44]

Given

R = \sqrt{d^2 + z^2}

and

\cos \alpha = \frac{d}{R}

find

\int \frac{1}{d^2 + z^2}dz

The answer given is

\frac{1}{d}\tan^{-1}(\frac{z}{d})

I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.

Use a right triangle to get the relationships in your trig substitution. In my triangle, the acute angle a (for alpha) is to the left, and the right angle is at the right. The base of the triangle is d and the altitude is z. The hypotenuse is sqrt(d^2 + z^2). You don't need R.

From this triangle, tan(a) = z/d, so d*tan(a) = z, and d*sec^2(a)da = dz. From the triangle you can also see that sec(a) = sart(d^2 + z^2)/d.

Make the appropriate substutions in your original integral, and find the antiderivative. When you are done, make sure you undo your substitution to get an antiderivative in terms of z.

 

[geft]

@dextercioby: It's not exactly a problem, but a guided solution. The book just sort of skips right to the answer without working out the integration, which got me confused.

@Mark44: Many thanks!