The value $$I=$$∫$$0$$π$$/2$$ $$11+$$$$cot$$⁡ xdx is

Question

The value $$I=$$∫$$0$$π$$/2$$ $$11+$$$$cot$$⁡ xdx is

Infinity Learn · Accepted Answer

Let $$f(x)=11+$$$$cot$$⁡ $$x=$$$$sin$$⁡ xsin⁡ $$x+$$$$cos$$⁡ x,$$f($$π$$/2$$−$$x)=$$$$sin$$⁡$$($$π$$/2$$−$$x)$$$$sin$$⁡$$($$π$$/2$$−$$x)+$$$$cos$$⁡$$($$π$$/2$$−$$x)=$$$$cos$$⁡xcos⁡$$x+$$$$sin$$⁡x Considering $$2I=$$∫$$0$$π$$/2$$ $$f(x)dx+$$∫$$0$$π$$/2$$ $$f($$π$$/2$$−$$x)dx=$$∫$$0$$π$$/2$$ $$sin$$⁡xcos⁡$$x+$$$$sin$$⁡$$xdx+$$∫$$0$$π$$/2$$ $$cos$$⁡xcos⁡$$x+$$$$sin$$⁡$$xdx=$$∫$$0$$π$$/2$$ $$sin$$⁡ $$x+$$$$cos$$⁡ xcos⁡ $$x+$$$$sin$$⁡ $$xdx=$$∫$$0$$π$$/2$$ $$1$$⋅$$dx=$$π$$2So$$ I $$=$$ π$$/4$$

The value $I = \int_{0}^{π / 2} \frac{1}{1 + \cot x} d x$ is

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The value I=∫0π/2 11+cot⁡ xdx is

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The value $I = \int_{0}^{π / 2} \frac{1}{1 + \cot x} d x$ is