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The integral \small \int_{\frac{\pi }{4}}^{\frac{\pi }{12}}  \small \frac{8\cos 2x}{\left ( \tan x+\cot x \right )^{3}} dx  equals :

  • Option 1)

    \frac{15}{128}

  • Option 2)

    \frac{15}{64}

  • Option 3)

    \frac{13}{32}

  • Option 4)

    \frac{13}{256}

 

Answers (1)

best_answer

As we learnt in

Integrals for Trigonometric functions -

\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos x \right ) =sinx

\therefore \int sinxdx=-cosx+c

-

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 )

 

 \int_{\frac{\pi }{4}}^{\frac{\pi }{12}}\frac{8\cos 2x}{\left ( \tan x+\cot x \right )^{3}}\:dx

=\int_{\frac{\pi }{4}}^{\frac{\pi }{12}}\frac{8\cos 2x}{\left ( \frac{2}{\sin2x } \right )^{3}}\:dx

=\int_{\frac{\pi }{4}}^{\frac{\pi }{12}}\frac{8\cos 2x.\sin ^{3}2x\:dx}{8}=\frac{1}{8} \left[sin^{4}2x \right ]_{\pi/4}^{\pi/12}

\Rightarrow \frac{15}{128}


Option 1)

\frac{15}{128}

Correct option    

Option 2)

\frac{15}{64}

Incorrect option    

Option 3)

\frac{13}{32}

Incorrect option    

Option 4)

\frac{13}{256}

Incorrect option    

Posted by

Plabita

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