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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)

best_answer

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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