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Mathematics Question on Integrals of Some Particular Functions

∫0xlog⁡ (cot⁡ x + tan⁡t) dt\int_{0}^{x}{\log \,(\cot \,x\,+\,\tan t)\,dt} =

A

x log⁡ (sin⁡ x)x\,\log \,(\sin \,x)

B

−x log⁡ (sin⁡ x)-x\,\log \,(\sin \,x)

C

x log⁡ (cos x)x\,\log \,(cos\,x)

D

−x log⁡ (cos x)-x\,\log \,(cos\,x)

Answer

−x log⁡ (sin⁡ x)-x\,\log \,(\sin \,x)

Explanation

Solution

Let I=∫0xlog⁡ (cot⁡ x+tan⁡t) dtI=\int_{0}^{x}{\log \,(\cot \,x+\tan t)\,dt}
=∫0xlog⁡(cos⁡xsin⁡x+sin⁡tcos⁡t)dt=\int_{0}^{x}{\log \left( \frac{\cos x}{\sin x}+\frac{\sin t}{\cos t} \right)dt}
=∫0x[log⁡ cos⁡ (x−t)−log⁡ sin⁡ x−log⁡ cos⁡t]dt=\int_{0}^{x}{[\log \,\\{\cos \,(x-t)\\}-\log \,\sin \,x-\log \,\cos t]dt}
=∫0xlog⁡ cos⁡(x−x+t) dt=\int_{0}^{x}{\log \,\\{\cos (x-x+t)\\}\,dt} −∫0xlog⁡  sin⁡x dt−∫0xlog⁡cos⁡ tdt-\int_{0}^{x}{\log \,\,\sin x\,dt-\int_{0}^{x}{\log \cos \,tdt}}
=∫0xlog⁡ cos⁡t dt−[t log⁡ sin⁡ x]0x−∫0xlog⁡  cos⁡ t  dt=\int_{0}^{x}{\log \,\cos t\,dt-[t\,\log \,\sin \,x]_{0}^{x}-\int_{0}^{x}{\log \,\,\cos \,t\,\,dt}}
=−(x log⁡ sin⁡x)=-(x\,\log \,\sin x)