Integral of 1/(x^2+5)^2 using a tangent trig substitution.

In this trigonometric substitution integral, we compute the integral of 1/(x^2+5)^2 using a tangent trig substitution.

The key feature to recognize in this integral is that we have a variable squared plus a constant. This means we can make a trig substitution to create a tangent squared plus one. To get the trigonometric substitution done, we let x=sqrt(5)*tan(theta), so when we square the x, we can factor a 5 out of the denominator.

Next, we compute the differential of x in the trig substitution, and we get dx=sqrt(5)*sec^2(theta)d(theta). We transform the integral to theta space, factor two factors of 5 out of the denominator, move the constants outside the integral, and arrive at an integral of cos^2(theta)d(theta).

The integral of cos^2(theta) is a classic trigonometric integral, and it requires the trig identity cos^2(theta)=1/2*(1+cos(2*theta)). Making the substitution, we arrive at a theta integral with easily guessable pieces, so we get the antiderivative of 1 and the antiderivative of cos(2*theta). Unfortnately, we aren't done, because we need to express the antiderivative in terms of x.

To transform the antiderivative in terms of x, we have to substitute theta in terms of x as arctan(x/sqrt(5)). This leads to a problem, because one of the terms in our antiderivative is sin(2*theta), but we can't compute sin(2*arctan(x/sqrt(5)). We have to apply the identity sin(2x)=2cosx*sinx in order to arrive at an expression that has trig functions of inverse trig functions.

Now we have to compute sin(arctan(x/sqrt(5)) and cos(arctan(x/sqrt(5)), and we do this by visualizing the angle whose tangent is x/sqrt(5) with a right triangle. The opposite side is x and the adjacent side is sqrt(5), and this means the hypotenuse is sqrt(x^2+5). Now we can compute the sine and cosine of the angle and complete the conversion of the antiderivative in terms of x to get the final result of the integral.

Note that this type of integral (an integral of one over the square of an irreducible quadratic) occurs commonly with integration by partial fractions decomposition, so it's a good type of integral to know!