1.

Compute \(\rm \int {2\sin x\over \cos^2x + \cos x -6 }\) dx1.  \(\rm {2}\left[ln(\cos x+3)+ln(2-\cos x)\right]\) + C2.  \(\rm {2\over5}\left[\ln(\cos x+3)+\ln(2-\cos x)\right]\) + C3.  \(\rm {2}\left[\ln(\cos x+3)-\ln(\cos x-2)\right]\) + C4.  \(\rm {2\over5}\left[\ln(\cos x+3)-\ln(2-\cos x)\right]\) + C

Answer» Correct Answer - Option 4 :  \(\rm {2\over5}\left[\ln(\cos x+3)-\ln(2-\cos x)\right]\) + C

Concept:

Integral property:

  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C
  • ∫ ex dx = ex+ C
  • ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C

 

 

Calculation:

I = \(\rm \int {2\sin x\over \cos^2x + \cos x -6 }\) dx

⇒ I = \(\rm -2\int {(-\sin x)\over \cos^2x + \cos x -6 }\) dx

By substitution let cos x = t ⇒ (- sin x) dx = dt

⇒ I = \(\rm -2\int {1\over t^2 + t -6}\) dt

⇒ I = \(\rm -2\int {1\over (t-2)(t+3)}\) dt

⇒ I =  \(\rm -2\int {1\over5}\left[{1\over (t-2)}-{1\over(t+3)}\right]\) dt

⇒ I = \(\rm {2\over5}\int\left[{1\over (t+3)}-{1\over(t-2)}\right] dt\)

⇒ I = \(\rm {2\over5}\left[\ln|t+3|-\ln|t-2|\right]\) + C

⇒ I = \(\rm {2\over5}\left[\ln|\cos x+3|-\ln|\cos x-2|\right]\) + C

∵ -1 ≤ cos x ≤ 1 

⇒ I = \(\boldsymbol{\rm {2\over5}\left[\ln(\cos x+3)-\ln(2-\cos x)\right]}\) + C


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