\(\int \left(\frac{cot^22x - 1}{2cot 2x} - cos \,8x \,cot\,4x\right)dx\)
\(= \frac12 \int cot\, 2x \;dx - \frac12\int tan\,2x\;dx - \int (1 - 2 sin^24x) \frac{cos\,4x}{sin\,4x}dx\)
\( = \frac12 \frac{log\; sin\,2x}2- \frac12\frac{log\;sec\,2x }2 - \int(\frac1{sin\,4x}- 2sin\,4x) cos\,4x\;dx\)
Let
\(sin\,4x = t\)
\(4\,cos\, 4x \;dx = dt\)
⇒ \(cos\,4x\;dx = \frac{dt}4\)
\( = \frac14 \;log\;sin\,2x - \frac14 \;log\;sec\,2x - \frac14 \int \left(\frac1t- 2t\right)dt\)
\(= \frac14 \;log\left(\frac{sin\,2x}{sec\,2x}\right) - \frac14 (log\, t - t^2) + C\)
\(= \frac 14 \;log(sin\,2x.cos\,2x)-\frac14 \;log \, t + \frac14 t^2 + C\)
\( = \frac14 \;log \left(\frac{sin\,4x}{2}\right) - \frac14 \;log (sin\,4x)+ \frac{sin^2\, 4x}{4} + C\) \((\because t = sin\,4x)\)
\(= \frac14 \; log\,\left(\frac{sin\,4x }{2}\times \frac1 {sin\,4x}\right)+ \frac{sin^24x}{4}+ C\)
\(= -\frac14\; log\, 2 + \frac{sin^24x}{4}+ C\)
\(= \frac{sin^2 4x}{4}+ k, \text{where} \;k = C - \frac {log\,2}4\)
