The integral ∫ π 12 π 4 8 cos 2 x tan x + cot x 3 d x equals

Given ∫ π 12 π 4 8 cos 2 x tan x + cot x 3 d x = ∫ π 12 π 4 8 cos 2 x sin x cos x + cos x sin x 3 d x = ∫ π 12 π 4 cos 2 x 1 sin 2 x 3 = ∫ π 12 π 4 cos 2 x · sin 2 x · sin 2 2 x d x = 1 4 ∫ π 12 π 4 sin 4 x · ( 1 − cos 4 x ) d x = 1 4 ∫ π 12 π 4 sin 4 x - 1 8 ∫ π 12 π 4 sin 8 x = − 1 16 cos 4 x π 12 π 4 + 1 8 × 8 cos 8 x π 12 π 4 = − 1 16 − 1 − 1 2 + 1 64 1 + 1 2 = 15 128 .

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationMedium2 minPYQ_2017

MathematicsMediumsingle choice

The integral $\int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 \cos 2 x}{{(\tan x + \cot x)}^{3}} d x$ equals

Options:

Answer:

Solution:

Given

$\int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 \cos 2 x}{{(\tan x + \cot x)}^{3}} d x = \int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 \cos 2 x}{{(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x})}^{3}} d x$

$= \int_{\frac{π}{12}}^{\frac{π}{4}} \frac{\cos 2 x}{{(\frac{1}{\sin 2 x})}^{3}} = \int_{\frac{π}{12}}^{\frac{π}{4}} \cos 2 x \cdot \sin 2 x \cdot \sin^{2} (2 x) d x$

$= \frac{1}{4} \int_{\frac{π}{12}}^{\frac{π}{4}} \sin 4 x \cdot (1 - \cos 4 x) d x$

$= \frac{1}{4} \int_{\frac{π}{12}}^{\frac{π}{4}} \sin 4 x - \frac{1}{8} \int_{\frac{π}{12}}^{\frac{π}{4}} \sin 8 x$

$= - \frac{1}{16} {[\cos 4 x]}_{\frac{π}{12}}^{\frac{π}{4}} + \frac{1}{8 \times 8} {[\cos 8 x]}_{\frac{π}{12}}^{\frac{π}{4}}$

$= - \frac{1}{16} [- 1 - \frac{1}{2}] + \frac{1}{64} [1 + \frac{1}{2}]$

$= \frac{15}{128}$ .

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: PYQ_2017

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