If I n = ∫ π 4 π 2 cot n x d x , then

I n = ∫ π / 4 π / 2 cot n x d x = ∫ π / 4 π / 2 cot n - 2 x cosec 2 x - 1 d x = - cot n - 1 x n - 1 π / 4 π / 2 - I n - 2 = 1 n - 1 - I n - 2 ⇒ I n + I n - 2 = 1 n - 1 ⇒ I 2 + I 4 = 1 3 I 3 + I 5 = 1 4 I 4 + I 6 = 1 5 ∴ 1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in A . P .

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDefinite Integration by PartsMedium2 minPYQ_2021

MathematicsMediumsingle choice

If $I_{n} = \int_{\frac{π}{4}}^{\frac{π}{2}} \cot^{n} x d x$ , then

Options:

Answer:

Solution:

$I_{n} = \int_{π / 4}^{π / 2} \cot^{n} x d x = \int_{π / 4}^{π / 2} \cot^{n - 2} x ({cosec}^{2} x - 1) d x$

${= - \frac{\cot^{n - 1} x}{n - 1}]}_{π / 4}^{π / 2} - I_{n - 2}$

$= \frac{1}{n - 1} - I_{n - 2}$

$\Rightarrow I_{n} + I_{n - 2} = \frac{1}{n - 1}$

$\Rightarrow I_{2} + I_{4} = \frac{1}{3}$

$I_{3} + I_{5} = \frac{1}{4}$

$I_{4} + I_{6} = \frac{1}{5}$

$∴ \frac{1}{I_{2} + I_{4}}, \frac{1}{I_{3} + I_{5}}, \frac{1}{I_{4} + I_{6}}$ are in $A . P .$

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

⏱ 2mℹ️ Source: PYQ_2021

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