∫$$0$$π log⁡$$(1+$$$$cos$$⁡$$x)dx$$ is equal to

Question

∫$$0$$π log⁡$$(1+$$$$cos$$⁡$$x)dx$$ is equal to

Infinity Learn · Accepted Answer

Let,$$I=$$∫$$0$$π log⁡(1+$$$$cos$$⁡$$x)dx----iI=$$∫$$0$$π log⁡$${1+$$$$cos$$⁡$$($$π−$$x)}dx=$$∫$$0$$π log⁡$$(1$$−$$cos$$⁡$$x)dx$$[∵$$cos$$⁡$$($$π−$$x)=$$−$$cos$$⁡x]… $$(ii) $$=$$∫$$0$$π log⁡$$2sin2$$⁡$$x2dx$$ ∵$$1$$−$$cos$$⁡$$x=2sin2$$⁡$$x2=$$∫$$0$$π log⁡$$2+2log$$⁡$$sin$$⁡$$x2dx=$$∫$$0$$π log⁡$$2dx+2$$∫$$0$$π log⁡$$sin$$⁡$$x2dxIn$$ the second. integral, put  $$x2=t$$⇒$$dx=2dtand$$ limits when x $$=0$$, t $$=0$$ and when $$x=$$π,$$t=$$π$$/2$$∴ $$I=log$$⁡$$2$$[x]$$0$$π$$+2$$∫$$0$$π$$/2$$ log⁡$$($$$$sin$$⁡$$t)2dt=(log$$⁡$$2)($$π−$$0)+4$$−π$$2log$$⁡$$2$$∵∫$$0$$π$$/2$$ log⁡$$sin$$⁡$$xdx=$$−π$$2log$$⁡$$2=$$−πlog⁡$$2$$

$\int_{0}^{π} \log (1 + \cos x) dx$ is equal to

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∫0π log⁡(1+cos⁡x)dx is equal to

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$\int_{0}^{π} \log (1 + \cos x) dx$ is equal to