First, to prevent loss in presentation marks, we always tag on the dx on the end of the integral. That is, our goal is to find ∫(tan8x)dx.
Notice that tan8x=tan6xtan2x=tan6x(sec2x−1)=tan6xsec2x−tan6x
Then ∫(tan8x)dx=∫(tan6x)(sec2x)dx−∫(tan6x)dx
To solve the first integral, we let u=tanx which implies that du=sec2xdx, thus the integral is simplified to
∫u6du=u77+K=tan7x7+K.
Using similar methods to find ∫(tan6x)dx, ∫(tan4x)dx and ∫(tan2x)dx, we derive that
∫(tan8x)dx=tan7x7−tan5x5+tan3x3−tanx+x+K
For a slightly shorter method, we use the method of reduction formulae. Consider the general integral In=∫(tannx)dx. Using the same trigonometric identity, we find that
∫(tannx)dx=∫(tann−2x)(sec2x)dx−∫(tann−2x)dx
which stands to reason that In=tann−1xn−1−In−2.
All we now need to do is find I8, which in turn requires us to find I6, I4, I2 and I0 respectively.
I0 is fairly easy, since ∫(tan0x)dx=∫1dx=x+K.
Using the formulae given (In=tann−1xn−1−In−2), we can find everything else.
I2=tanx−I0=tanx−x−K
I4=tan3x3−I2=tan3x3−tanx+x+K
I6=tan5x5−I4=tan5x5−tan3x3+tanx−x−K
I8=tan7x7−I6=tan7x7−tan5x5+tan3x3−tanx+x+K
Therefore, ∫(tan8x)dx=tan7x7−tan5x5+tan3x3−tanx+x+K
