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wat will be d integration of
tan(x-k).tan(x+k).tan2xdx.......???
tan2x=tan[(x-k)+(x+k)]
→tan2x=[tan(x-k)+tan(x+k)]/[1-tan(x-k).tan(x+k)]
cross multiplying,we get
tan2x-tan2x.tan(x-k).tan(x+k)=tan(x-k)+tan(x+k)
→tan2x.tan(x-k).tan(x+k)=tan2x-tan(x-k)-tan(x+k)...............(1)
now,
∫tan(x-k).tan(x+k).tan2xdx
=∫[tan2x-tan(x-k)-tan(x+k)]dx......................from(1)
=∫tan2xdx-∫tan(x-k)dx-∫tan(x+k)dx
=(1/2)lnιsec2xι-lnιsec(x-k)ι-lnιsec(x+k)ι+c......where ιxι represents modulus of x,
and convert the soln. further according to the logarithm rules.
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