The integral ∫ π 6 π 3 tan 3 x · sin 2 3 x 2 sec 2 x · sin 2 3 x + 3 tan x · sin 6 x d x is equal to:

Rearranging, we get, ∫ π / 6 π / 3 d d x tan 4 x 2 · sin 4 3 x + tan 4 x · d d x sin 4 3 x 2 d x = 1 2 ∫ π / 6 π / 3 d d x tan 4 x · sin 4 3 x d x = 1 2 tan 4 x · sin 4 3 x π / 6 π / 3 = 1 2 3 4 × 0 - 1 3 4 × 1 = - 1 2 × 1 9 = - 1 18

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDefinite Integration by SubstitutionMedium2 minPYQ_2020

MathematicsMediumsingle choice

The integral $\int_{\frac{π}{6}}^{\frac{π}{3}} \tan^{3} x \cdot \sin^{2} 3 x (2 \sec^{2} x \cdot \sin^{2} 3 x + 3 \tan x \cdot \sin 6 x) d x$ is equal to:

Options:

Answer:

Solution:

Rearranging, we get,

$\int_{π / 6}^{π / 3} (\frac{\frac{d}{d x} (\tan^{4} x)}{2} \cdot \sin^{4} 3 x + \tan^{4} x \cdot \frac{\frac{d}{d x} (\sin^{4} 3 x)}{2}) d x$

$= \frac{1}{2} \int_{π / 6}^{π / 3} (\frac{d}{d x} (\tan^{4} x \cdot \sin^{4} 3 x)) d x$

$= \frac{1}{2} {[\tan^{4} x \cdot \sin^{4} 3 x]}_{π / 6}^{π / 3}$

$= \frac{1}{2} [{(\sqrt{3})}^{4} \times 0 - {(\frac{1}{\sqrt{3}})}^{4} \times 1]$

$= - \frac{1}{2} \times \frac{1}{9} = - \frac{1}{18}$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Definite Integration by Substitution

⏱ 2mℹ️ Source: PYQ_2020

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