Q21: Evaluate $\int \frac{dx}{x\sqrt{x^6 - 1}}$⁠, or $\int \frac{\cos x - \sin x}{\sqrt{\sin 2x}}\,dx$

We multiply and divide by $6x^5$⁠. This does not change the integrand:

Now $6x^5\,dx$ sits in the numerator. It is the differential of $x^6 - 1$⁠, so the substitution in the next line can swallow it.

Differentiate both sides of $x^6 - 1 = t^2$⁠: the left side gives $6x^5\,dx$⁠, the right side gives $2t\,dt$⁠.

Putting the root equal to $t$ (that is, $x^6 - 1 = t^2$⁠) is the trick: then $\sqrt{x^6 - 1} = t$⁠, and the square root disappears.

The same equation gives $x^6 = t^2 + 1$ for the $x^6$ left in the denominator.

Replace each piece: $6x^5\,dx$ by $2t\,dt$⁠, $x^6$ by $t^2 + 1$⁠, and $\sqrt{x^6 - 1}$ by $t$⁠:

The $t$ cancels, and $\frac16 \times 2 = \frac13$⁠.

Then use the standard integral

The question is in $x$⁠, so the answer must be in $x$⁠. From $x^6 - 1 = t^2$⁠, $t = \sqrt{x^6 - 1}$⁠.

Watch out: do not stop at $\frac13\tan^{-1}t + C$⁠. Leaving the answer in $t$ loses the last mark.

The numerator $\cos x - \sin x$ is the derivative of $\sin x + \cos x$⁠. So we want $\sin 2x$ written in terms of $\sin x + \cos x$⁠.

Squaring and using $\sin^2x + \cos^2x = 1$⁠:

The last line uses the double angle identity $2\sin x\cos x = \sin 2x$⁠.

Subtract 1 from both sides to get $\sin 2x$⁠.

Differentiate: $\frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$⁠.

So the whole numerator, together with $dx$⁠, becomes $dt$⁠, and the root becomes $\sqrt{t^2 - 1}$⁠.

This is the standard integral with $a = 1$⁠:

Watch out: it is not $\sin^{-1}$⁠. That one belongs to $\sqrt{a^2 - x^2}$⁠, where the square comes second.

Replace $t$ and $t^2 - 1$ one at a time: