00:01
To integrate this, note that we can rewrite this into the integral of 1 over, we have x squared plus the square of square root of 5 raised to 3 over 2 dx.
00:13
Note that this follows the format of the radical x squared plus a squared, which uses x equals a tangent theta as its trigonometric substitution.
00:27
So in here we let x equal to a which is square to 5 tangent theta.
00:35
And the differential of x would be dx, it's equal to squared of 5, secan squared theta, d theta.
00:42
So using this, we have the integral of 1 over, we have x squared, that's square of 5 tangent theta squared, plus square to 5 square, that's just 5, raised to 3 over 2, and dx would be, squared of 5, sequins squared theta, d theta.
01:01
So simplifying this, we have the integral of squared of 5, sequenced squared theta d theta, all over, we have 5 tangent squared theta plus 5, this thing raised to 3 over 2, and factoring out the 5 inside this parenthesis, we have the integral of square of 5, secan squared theta over we have 5 times tangent squared theta plus 1 this raise to 3 over 2 and so we have the integral of squared of 5 times secan squared theta over 5 raised to 3 halves and then times tangent squared theta plus 1 which is just secan squared theta so we have sicken squared theta raised to 3 over 2 and then d theta and then from here we can cancel that squared of five so we have one here and it should be five in here and then in here this should simplify into secant raised to the third power and canceling out a secund squared would be one over secant theta left and so we have from here the integral of one 1 over 5 secan theta, d theta, which is the same as 1 over 5 integral of cosine theta d theta.
02:41
Now the antiderivative of cosine is sine, and so we have 1 over 5 sine theta plus c...