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\int \frac{dx}{5+4\cos x}= K\tan^{-1}\left ( M\tan \left ( \frac{x}{2} \right ) \right )+C

  • Option 1)

    K = 1, M = 2/3

  • Option 2)

    K = 2/3, M = 1/3

  • Option 3)

    K = 1, M = 1/3

  • Option 4)

    K=1, M = 1/3

 

Answers (1)

As we learnt

Special type of indefinite integration -

Integration of the form : g(sin^{-1}f(x))

Put f(x)=t 

so f'(x)dx=dt

Now Put \Theta =sin^{-1}t

such that t=sin\Theta

-

 

 

Since the integrand is a rational function of cos x, we put tan(x/2) = t.

                              Then

I=\int {\frac{{dx}}{{5 + 4\cos \,x}}}=\int {\frac{{2dt}}{{\left( {1 + {t^2}} \right)\left( {5 + \frac{{4\left( {1 - {t^2}} \right)}}{{1 + {t^2}}}} \right)}}} $=2\int {\frac{{dt}}{{5\left( {1 + {t^2}} \right) + 4\left( {1 - {t^2}} \right)}}} $

= 2\int {\frac{{dt}}{{{t^2} + 9}} = \frac{2}{3}} $ta{n^{ - 1}}\frac{t}{3}\; + {\rm{ }}C{\rm{ }} = \frac{2}{3}\;ta{n^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right)$+C

Hence K = 2/3 and M = 1/3.


Option 1)

K = 1, M = 2/3

Option 2)

K = 2/3, M = 1/3

Option 3)

K = 1, M = 1/3

Option 4)

K=1, M = 1/3

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