If ∫$$sec4/3$$⁡$$xcosec8/3$$⁡$$xdx=a($$$$tan$$⁡$$x)$$−$$5/3+b($$$$tan$$⁡$$x)1/3+C$$, then $$5a+b=$$

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If ∫$$sec4/3$$⁡$$xcosec8/3$$⁡$$xdx=a($$$$tan$$⁡$$x)$$−$$5/3+b($$$$tan$$⁡$$x)1/3+C$$, then $$5a+b=$$

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We have,          I∫$$sec4/3$$⁡$$xcosec8/3$$⁡xdx⇒ $$I=$$∫$$cos$$−$$4/3$$⁡xsin−$$8/3$$⁡$$xdx=$$∫$$1cos4/3$$⁡$$xsin8/3$$⁡xdxThe sum of the exponents of $$sin$$ x and $$cos$$ x is $$-$$ $$4$$, an even integer. So, we divide both numerator and denominator by $$cos4x$$.∴ I    $$=$$∫$$sec4$$⁡$$xtan8/3$$⁡$$xdx=$$∫$$1+tan2$$⁡$$x($$$$tan$$⁡$$x)$$−$$8/3d($$$$tan$$⁡$$x)$$⇒ I    $$=$$∫$$($$$$tan$$⁡$$x)$$−$$8/3+($$$$tan$$⁡$$x)$$−$$2/3d($$$$tan$$⁡$$x)$$⇒ I    $$=$$−$$35($$$$tan$$⁡$$x)$$−$$5/3+3($$$$tan$$⁡$$x)1/3+C$$∴ $$a=$$−$$35$$ and $$b=3$$⇒$$5a+b=0$$

If $\int \sec^{4 / 3} x {cosec}^{8 / 3} x d x = a (\tan x)^{- 5 / 3} + b (\tan x)^{1 / 3} + C,$ then $5 a + b =$

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If ∫sec4/3⁡xcosec8/3⁡xdx=a(tan⁡x)−5/3+b(tan⁡x)1/3+C, then 5a+b=

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If $\int \sec^{4 / 3} x {cosec}^{8 / 3} x d x = a (\tan x)^{- 5 / 3} + b (\tan x)^{1 / 3} + C,$ then $5 a + b =$