Integral Calculation Exercise 32

Let’s evaluate the following integral:

$$ \int \frac{1}{\tan^3(x)} \ dx $$

Since the tangent function is defined as the ratio of sine to cosine, tan = sin / cos, we can rewrite the integrand accordingly:

$$ \int \frac{1}{ \frac{\sin^3(x)}{\cos^3(x)} } \ dx $$

Using the rule for dividing fractions, we multiply the expression by $ \frac{\cos^3(x)}{\sin^3(x)} $ over itself to simplify:

$$ \int \frac{1}{ \frac{\sin^3(x)}{\cos^3(x)} } \cdot \frac{ \frac{\cos^3(x)}{\sin^3(x)} }{ \frac{\cos^3(x)}{\sin^3(x)} } \ dx $$

After simplifying, we’re left with:

$$ \int \frac{\cos^3(x)}{\sin^3(x)} \ dx $$

Now, apply the Pythagorean identity $ \cos^2(x) + \sin^2(x) = 1 $, which allows us to write:

$ \cos^2(x) = 1 - \sin^2(x) $

So we express the integrand as:

$$ \int \frac{\cos^2(x) \cdot \cos(x)}{\sin^3(x)} \ dx $$

$$ \int \frac{[1 - \sin^2(x)] \cdot \cos(x)}{\sin^3(x)} \ dx $$

At this point, we make the substitution $ u = \sin(x) $:

$$ u = \sin(x) $$

Then compute the differential:

$$ du = \cos(x) \ dx $$

Substituting into the integral, we obtain:

$$ \int \frac{(1 - u^2) \cdot \cos(x)}{u^3} \ dx $$

$$ \int \frac{1 - u^2}{u^3} \cdot \cos(x) \ dx $$

Now, substitute $ \cos(x) dx = du $:

$$ \int \frac{1 - u^2}{u^3} \ du $$

Split the fraction into two simpler terms:

$$ \int \left( \frac{1}{u^3} - \frac{u^2}{u^3} \right) \ du $$

$$ \int \left( \frac{1}{u^3} - \frac{1}{u} \right) \ du $$

Which becomes:

$$ \int u^{-3} \ du - \int \frac{1}{u} \ du $$

The second integral is a standard logarithmic form:

$ \int \frac{1}{u} \ du = \log |u| + C $

The first is a power function:

$ \int u^{-3} \ du = \frac{u^{-2}}{-2} + C $

Putting it all together:

$$ \frac{u^{-2}}{-2} - \log |u| + C $$

$$ -\frac{1}{2u^2} - \log |u| + C $$

Finally, substituting back $ u = \sin(x) $:

$$ -\frac{1}{2 \sin^2(x)} - \log | \sin(x) | + C $$

This is the final result of the integral.

And that concludes the calculation.