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The integral   \int_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\tan 2x}} is equal to :

  • Option 1)

    \frac{\pi }{18}

  • Option 2)

    \frac{\pi }{3}

  • Option 3)

    \frac{\pi }{12}

  • Option 4)

    \frac{\pi }{6}

 

Answers (1)

best_answer

As learnt in concept

Properties of Definite integration -

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

When \int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx

 

- wherein

Put the \left ( a+b-x \right ) at the place of x in f\left ( x \right )

 

 \int_{\frac{\pi }{24}}^{\frac{5\pi}{24}}\frac{dx}{1+\sqrt[3]{tan2x}}

Let \: 2x=t; dx=\frac{dt}{2}

I=\frac{1}{2}\int_{\frac{\pi }{12}}^{\frac{5\pi }{12}}\frac{dt}{1+(tan \:t)^\frac{1}{3}}

I=\frac{1}{2}\int_{\frac{\pi }{12}}^{\frac{5\pi }{12}}\frac{dt}{1+(cot \:t)^\frac{1}{3}} \left[\int_{a}^{b}f(x)=\int_{a}^{b}f(a+b-x) \right ]

2I=\frac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi }{12}}dt

I= \frac{1}{4}\times \frac{4\pi }{12} =\frac{\pi }{12}

 


Option 1)

\frac{\pi }{18}

Incorrect 

Option 2)

\frac{\pi }{3}

Incorrect 

Option 3)

\frac{\pi }{12}

Correct

Option 4)

\frac{\pi }{6}

Incorrect 

Posted by

Aadil

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