How do you find the antiderivative of #int sec^2xcsc^2x dx#?

Question

Ezekiel Alexander · Accepted Answer

#tanx - cotx + C#

Do a little bit of experimentation using some trig identities. Recall the pythagorean identity #color(red)(csc^2alpha = 1 + cot^2alpha)#.

#=intsec^2x(1 + cot^2x)dx#

Now recall that cotangent function is the reciprocal of the tangent function and the secant function is the reciprocal of the cosine function.

#=int1/cos^2x(1 + cos^2x/sin^2x)dx#

#=int 1/cos^2x + 1/sin^2xdx#

#=int sec^2x + csc^2xdx#

#=intsec^2x + intcsc^2xdx#

These are both widely known integrals. If you haven't already, I would recommend learning them by heart.

#=tanx - cotx + C#

Hopefully this helps!

Elijah Abram · Answer

undefined To find the antiderivative of $ \int \sec^2(x) \csc^2(x) \, dx $, we can use trigonometric identities to simplify the expression.

First, we can rewrite $ \sec^2(x) $ as $ \frac{1}{\cos^2(x)} $ and $ \csc^2(x) $ as $ \frac{1}{\sin^2(x)} $.

So, $ \int \sec^2(x) \csc^2(x) \, dx $ becomes $ \int \frac{1}{\cos^2(x)} \cdot \frac{1}{\sin^2(x)} \, dx $.

Next, we can use the identity $ \sin^2(x) + \cos^2(x) = 1 $ to rewrite $ \frac{1}{\cos^2(x)} $ as $ \frac{1}{1 - \sin^2(x)} $.

Substituting this into our integral, we get $ \int \frac{1}{1 - \sin^2(x)} \cdot \frac{1}{\sin^2(x)} \, dx $.

We can now let $ u = \sin(x) $, then $ du = \cos(x) \, dx $.

Substituting $ u $ and $ du $, the integral becomes $ \int \frac{1}{1 - u^2} \cdot \frac{1}{u^2} \, du $.

This integral can be solved by partial fraction decomposition. After finding the partial fractions, we integrate each term separately.

After integrating, we can substitute $ u = \sin(x) $ back into the result to get the antiderivative in terms of $ x $.