Question:

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

Updated On: Jul 27, 2024
  • x log⁡ (sin⁡ x) x\,\log \,(\sin \,x)
  • −x log⁡ (sin⁡ x) -x\,\log \,(\sin \,x)
  • x log⁡ (cos x) x\,\log \,(cos\,x)
  • −x log⁡ (cos x) -x\,\log \,(cos\,x)
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The Correct Option is B

Solution and Explanation

Let I=∫0xlog⁡ (cot⁡ x+tan⁡t) dt I=\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)
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Concepts Used:

Integrals of Some Particular Functions

There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.

Integrals of Some Particular Functions:

  • ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
  • ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
  • ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
  • ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
  • ∫1/√(a2 – x2) dx = sin-1(x/a) + C
  • ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C

These are tabulated below along with the meaning of each part.