In - -01 xn tan-1 x dx- If an In-2 - bn In - cn - n - n - 1 then

I_n = ( x^n+1/n+1 tan^ 1x )^1_0 ∫_0^1x^n+1/n+1.1/1+x^2dx (n+1)I_n = π/4 ∫_0^1x^n+1/1+x^2 dx (n+3)I_n+2 = π/4 ∫_0^1x^n+3/1+x^2dx ∴ (n+1) I_n + (n+3)I_n+ ...

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Byju's Answer

Standard XIII

Mathematics

Integration by Parts

In = ∫01 xn t...

Question

$I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1$ then

a_{1}, a_{2}, a_{3}

.....are in A.P.

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b_{1}, b_{2}, b_{3}

.....are in G.P.

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c_{1}, c_{2}, c_{3}

.....are in H.P.

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a_{1}, a_{2}, a_{3}

.....are in H.P.

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Solution

The correct option is A $a_{1}, a_{2}, a_{3}$ .....are in A.P.
$I_{n} = {(\frac{x^{n + 1}}{n + 1} t a n^{- 1} x)}_{0}^{- 1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{1}{1 + x^{2}} d x (n + 1) I_{n} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^{2}} d x (n + 3) I_{n + 2} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 3}}{1 + x^{2}} d x ∴ (n + 1) I_{n} + (n + 3) I_{n + 2} = \frac{π}{2} - \frac{1}{n + 2} ∴ a_{n} = (n + 3) \Rightarrow a_{1}, a_{2}, a_{3}$ .....are in A.P.
$b_{n} = (n + 1) \Rightarrow b_{1}, b_{2}$ ..... are in A.P.
$c_{n} = \frac{π}{2} - \frac{1}{n + 2}$ not in any progression.

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Similar questions

Q. Let

I_{n} = 1 \int 0 x^{n} {tan}^{- 1} x d x .

a_{n} I_{n + 2} + b_{n} I_{n} = c_{n}

for all

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I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1

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Q. If

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In=∫10xntan−1xdx. If anIn+2+bnIn=cn ∀ nϵN,n≥1 then

The correct option is A a1,a2,a3.....are in A.P.In=(xn+1n+1tan−1x)10−∫10xn+1n+1.11+x2dx(n+1)In=π4−∫10xn+11+x2dx(n+3)In+2=π4−∫10xn+31+x2dx∴(n+1)In+(n+3)In+2=π2−1n+2∴an=(n+3)⇒a1,a2,a3.....are in A.P. bn=(n+1)⇒b1,b2..... are in A.P. cn=π2−1n+2 not in any progression.

$I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1$ then