∫$$tan$$(x$$−α$$)$$tan$$$$(x+$$α$$)tan2xdx=$$

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Infinity Learn · Accepted Answer

$$I=$$∫$$tan$$⁡(x$$−α$$)$$tan$$⁡(x+$$α$$)$$tan$$⁡$$2xdx_$$ Now $$tan$$⁡$$2x=$$$$tan$$⁡(x+$$α$$+x$$−α$$)$$tan$$⁡$$2x=$$$$tan$$⁡(x+$$α$$)+$$$$tan$$⁡$$(x$$−α$$)1$$−$$tan$$⁡$$(x+$$α$$)$$tan$$⁡(x$$−α$$)$$tan$$⁡$$2x(1$$−$$tan$$⁡(x+$$α$$)$$tan$$⁡(x$$−α$$)=$$$$tan$$⁡$$(x+$$α$$)+$$$$tan$$⁡$$(x$$−α$$)∴$$tan$$⁡(x$$−α$$)$$tan$$⁡(x+$$α$$)$$tan$$⁡$$2x=$$$$tan$$⁡$$2x$$−$$tan$$⁡(x+$$α$$)−$$tan$$⁡(x$$−α$$)⇒$$I=$$∫$$tan$$⁡(x$$−α$$)$$tan$$⁡(x+$$α$$)$$tan$$⁡$$2xdx=$$∫$${$$$$tan$$⁡$$2x$$−$$tan$$⁡(x+$$α$$)−$$tan$$⁡(x$$−α$$)}dx=12log$$⁡|$$sec$$⁡$$2x$$|−log⁡|$$sec$$⁡$$(x+$$α$$)|−log⁡|$$sec$$⁡(x$$−α$$)|$$+C$$

$\int \tan (x - α) \tan (x + α) \tan 2 x d x =$

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∫tan(x−α)tan(x+α)tan2xdx=

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$\int \tan (x - α) \tan (x + α) \tan 2 x d x =$