The integral -245-24dx1-tan2x is equal to

Let I=∫_π/24^5π/24dx(1+√(tan2x)) ⇒ I=∫_π/24^5π/24√(cos2x)√(sin2x)+√(cos2x) dx ⋯(1) nbsp; Using property : ∫_a^bf(x) dx=∫_a^bf(a+b x) dx nbsp; I=∫_π/24^5π/ ...

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Byju's Answer

Standard VIII

Mathematics

Language of Ratio

The integral ...

Question

The integral $5 π / 24 \int π / 24 \frac{d x}{(1 + \sqrt[3]{tan 2 x})}$ is equal to

\frac{π}{3}

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\frac{π}{12}

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\frac{π}{6}

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\frac{π}{18}

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Solution

The correct option is B $\frac{π}{12}$
Let $I = 5 π / 24 \int π / 24 \frac{d x}{(1 + \sqrt[3]{tan 2 x})}$
$\Rightarrow I = 5 π / 24 \int π / 24 \frac{\sqrt[3]{cos 2 x}}{\sqrt[3]{sin 2 x} + \sqrt[3]{cos 2 x}} d x \dots (1)$
Using property : $b \int a f (x) d x = b \int a f (a + b - x) d x$
$I = 5 π / 24 \int π / 24 \frac{\sqrt[3]{sin 2 x}}{\sqrt[3]{sin 2 x} + \sqrt[3]{cos 2 x}} d x \dots (2)$
Adding $(1)$ and $(2),$ we get
$2 I = 5 π / 24 \int π / 24 \frac{\sqrt[3]{sin 2 x} + \sqrt[3]{cos 2 x}}{\sqrt[3]{sin 2 x} + \sqrt[3]{cos 2 x}} d x$
$\Rightarrow 2 I = 5 π / 24 \int π / 24 1 d x$
$\Rightarrow 2 I = \frac{5 π}{24} - \frac{π}{24}$
$∴ I = \frac{π}{12}$

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