The value of the definite integral $\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}}$ is :

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Definite Integrals

The value of the definite integral

\int_{π /24}^{5 π /24} \frac{d x}{1 + 3 t a n 2 x}

is :

Answer: (B)

Here

a = \frac{π}{24}

and

b = \frac{5 π}{24}

, so

a + b = \frac{π}{4}

and

2 (a + b - x) = \frac{π}{2} - 2 x

, which makes

tan (\frac{π}{2} - 2 x) = cot 2 x

. Applying

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

I = \int_{π /24}^{5 π /24} \frac{d x}{1 + 3 cot 2 x} = \int_{π /24}^{5 π /24} \frac{3 tan 2 x}{1 + 3 tan 2 x} d x

Adding this to the original form, the integrands combine to

1

2 I = \int_{π /24}^{5 π /24} d x = \frac{5 π}{24} - \frac{π}{24} = \frac{π}{6}

I = \frac{π}{12}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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