If b n = ∫ 0 π 2 cos 2 n x sin x d x , n ∈ ℕ , then

Given, b n = ∫ 0 π 2 1 + cos 2 n x sin x d x b n + 1 - b n = ∫ 0 π 2 cos 2 n + 1 x - cos 2 n x sin x d x = ∫ 0 π 2 - sin 2 n + 1 x sin x sin x d x = cos ( 2 n + 1 ) x 2 n + 1 0 π 2 = - 1 2 n + 1 So, 1 b 3 - b 2 = - 5 1 b 4 - b 3 = - 7 1 b 5 - b 4 = - 9 So, 1 b 3 - b 2 , 1 b 4 - b 3 , 1 b 5 - b 4 are in A.P. with c . d = - 2

Home

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDefinite Integration by SubstitutionMedium2 minPYQ_2022

MathematicsMediumsingle choice

If $b_{n} = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} n x}{\sin x} d x, n \in ℕ$ , then

Options:

Answer:

Solution:

Given,

$b_{n} = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 n x}{\sin x} d x$

$b_{n + 1} - b_{n} = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} (n + 1) x - \cos^{2} n x}{\sin x} d x$

$= \int_{0}^{\frac{π}{2}} \frac{- \sin (2 n + 1) x \sin x}{\sin x} d x$

$= {(\frac{\cos (2 n + 1) x}{2 n + 1})}_{0}^{\frac{π}{2}} = \frac{- 1}{2 n + 1}$

So, $\frac{1}{b_{3} - b_{2}} = - 5$

$\frac{1}{b_{4} - b_{3}} = - 7$

$\frac{1}{b_{5} - b_{4}} = - 9$

So, $\frac{1}{b_{3} - b_{2}}, \frac{1}{b_{4} - b_{3}}, \frac{1}{b_{5} - b_{4}}$ are in A.P. with $c . d = - 2$

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

⏱ 2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...