The integral \int_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{dx}{\sin 2x (\tan^{5}x + \cot ^{5}x)}  equals :

  • Option 1)

    \frac{1}{10}(\frac{\pi}{4}-\tan^{-1}(\frac{1}{9\sqrt3}))

  • Option 2)

    \frac{\pi}{40}

  • Option 3)

    \frac{1}{20}\tan^{-1}(\frac{1}{9\sqrt3})

  • Option 4)

    \frac{1}{5}(\frac{\pi}{4}-\tan^{-1}(\frac{1}{3\sqrt3}))

Answers (1)
A admin

 

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

Integration of Rational and irrational function -

Integration in the form of : 

(i) \int \frac{dx}{x^{2}+a^{2}} (ii) \int \frac{dx}{x^{2}-a^{2}} (iii) \int \frac{dx}{a^{2}-x^{2}}

(iv) \int \frac{dx}{\sqrt{x^{2}+a^{2}}} (v) \frac{dx}{\sqrt{x^{2}-a^{2}}} (vi) \frac{dx}{\sqrt{a^{2}-x^{2}}}

- wherein

(i)    \frac{1}{a}tan^{-1}\, \frac{x}{a}+C

(ii)    \frac{1}{2a}log \left | \frac{x-a}{x+a} \right |+C

(iii)    \frac{1}{2a}log \left | \frac{x+a}{x-a} \right |+C

(iv)    log\left | x+ \sqrt{x^{2}+a^{2}} \right |+C

(v)    log\left | x+ \sqrt{x^{2}-a^{2}} \right |+C

(vi)    sin^{-1} \left(\frac{x}{a} \right )+C

 

  I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{dx}{\sin 2x(\tan ^{5}x+\cot ^{5}x)}

    =\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{\tan ^{4}x\cdot \sec ^{2}xdx}{(\tan ^{10}x+1)}

Put  \tan ^{5}x=t

dt=5\tan ^{4}x\sec ^{2}x\: \: dx

I=\frac{1}{10}\int_{(\frac{1}{\sqrt3})^{5}}^{1}\frac{dt}{1+t^{2}}

    =\frac{1}{10}\left [ \frac{\pi}{4}-\tan^{-1}(\frac{1}{9\sqrt3}) \right ]

 


Option 1)

\frac{1}{10}(\frac{\pi}{4}-\tan^{-1}(\frac{1}{9\sqrt3}))

Option 2)

\frac{\pi}{40}

Option 3)

\frac{1}{20}\tan^{-1}(\frac{1}{9\sqrt3})

Option 4)

\frac{1}{5}(\frac{\pi}{4}-\tan^{-1}(\frac{1}{3\sqrt3}))

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