∫$$01$$ $$sin$$⁡$$2tan$$−$$1$$⁡$$1+x1$$−xdx is equal to

Question

∫$$01$$ $$sin$$⁡$$2tan$$−$$1$$⁡$$1+x1$$−xdx is equal to

Infinity Learn · Accepted Answer

Let , $$I=$$∫$$01$$ $$sin$$⁡$$2tan$$−$$1$$⁡$$1+x1$$−xdxput, $$x=$$$$cos$$⁡$$2$$θThen, $$sin$$⁡$$2tan$$−$$1$$⁡$$1+$$$$cos$$⁡$$2$$θ$$1$$−$$cos$$⁡$$2$$θ$$=$$$$sin$$⁡$$2tan$$−$$1$$⁡$$($$$$cot$$⁡θ$$)=$$$$sin$$⁡$$2tan$$−$$1$$⁡$$tan$$⁡π$$2$$−θ$$=$$$$sin$$⁡$$2$$$$π$$$$2$$−θ$$=$$$$sin$$⁡$$($$π−$$2$$θ$$)=$$$$sin$$⁡$$2$$θ$$=1$$−$$cos2$$⁡$$2$$θ$$=1$$−$$x2Now$$,∫$$01$$ $$sin$$⁡(2tan$$−$$1$$⁡$$1+x1$$−$$xdx=$$∫$$01$$ $$1$$−$$x2dx=12x$$⋅$$1$$−$$x201+12sin$$−$$1$$⁡$$x01=12$$[$$11$$−$$1$$−$$0$$]$$+12sin$$−$$1$$⁡$$(1)−$$0=12$$[$$0$$]$$+12$$$$π$$$$2=$$π$$4$$

$\int_{0}^{1} \sin (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) dx$ is equal to

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∫01 sin⁡2tan−1⁡1+x1−xdx is equal to

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$\int_{0}^{1} \sin (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) dx$ is equal to