Let - x-1-x-x2 dx-1-x-x23-2-x-12-1-x-x2-sin-12x-1-5-C where - - - are constants- Then the value of 3- iswhere C is integration constant

Let I=∫ x√(1+x x^2)dx⋯(1) Now, x=λddx(1+x x^2)+μ ⇒λ= 12, μ=12 So, x= 12(1 2x)+12⋯(2) From (1) and (2) I=∫[ 12(1 2x)+12]√(1+x x^2) dx =∫ 12(1 2x)(√(1+x x^2))d ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Quadratic Equation

Let ∫ x√1+x-x...

Question

Let $\int x \sqrt{1 + x - x^{2}} d x = α (1 + x - x^{2})^{3 / 2} + β [(x - \frac{1}{2}) \sqrt{1 + x - x^{2}} + γ {sin}^{- 1} \frac{2 x - 1}{\sqrt{5}}] + C$ (where $α, β, γ$ are constants). Then the value of $3 (\frac{β - γ}{α})$ is
$($ where $C$ is integration constant $)$

Open in App

Solution

Let $I = \int x \sqrt{1 + x - x^{2}} d x \dots (1)$
Now,
$x = λ \frac{d}{d x} (1 + x - x^{2}) + μ \Rightarrow λ = \frac{- 1}{2}, μ = \frac{1}{2}$
So,
$x = \frac{- 1}{2} (1 - 2 x) + \frac{1}{2} \dots (2)$
From $(1)$ and $(2)$
$I = \int [\frac{- 1}{2} (1 - 2 x) + \frac{1}{2}] \sqrt{1 + x - x^{2}} d x = \int \frac{- 1}{2} (1 - 2 x) (\sqrt{1 + x - x^{2}}) d x + \int \frac{1}{2} \sqrt{1 + x - x^{2}} d x$
Now,
$I_{1} = \int \frac{- 1}{2} (1 - 2 x) (\sqrt{1 + x - x^{2}}) d x = \frac{- 1}{3} (1 + x - x^{2})^{3 / 2} + C I_{2} = \int \frac{1}{2} \sqrt{1 + x - x^{2}} d x = \frac{1}{2} \int \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = \frac{1}{2} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1}{2} (x - \frac{1}{2}) \sqrt{(1 + x - x^{2})} + \frac{1}{2} {(\frac{\sqrt{5}}{2})}^{2} {sin}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + C = \frac{1}{4} [(x - \frac{1}{2}) \sqrt{(1 + x - x^{2})} + (\frac{5}{4}) {sin}^{- 1} \frac{2 x - 1}{\sqrt{5}}] + C \Rightarrow I = \frac{- 1}{3} (1 + x - x^{2})^{3 / 2} + \frac{1}{4} [(x - \frac{1}{2}) \sqrt{(1 + x - x^{2})} + (\frac{5}{4}) {sin}^{- 1} \frac{2 x - 1}{\sqrt{5}}] + C$
Here,
$α = \frac{- 1}{3}, β = \frac{1}{4}, γ = \frac{5}{4} ∴ 3 (\frac{β - γ}{α}) = 9$

Suggest Corrections

Similar questions

Q. Evaluate

\frac{(x + α) (x + β)}{(α - γ) (β - γ)} + \frac{(x + β) (x + γ)}{(β - α) (γ - α)} + \frac{(x + α) (x + γ)}{(γ - β) (α - β)} = ?

Q. In the given figure, x = ?
(a) α + β − γ
(b) α − β + γ
(c) α + β + γ
(d) α + γ − β

Q. If the coeffecient of the middle term in the expansion of

{(1 + x)}^{2 n + 2}

α

and coeffecient of middle terms in the expansion of

{(1 + x)}^{2 n + 1}

are

β

and

γ

, then relate

α, β

and

γ

Q. If the coefficient of the middle term in the expansion of

(1 + x)^{2 n + 2}

α

and the coefficients of middle terms in the expansion of

(1 + x)^{2 n + 1}

are

β

and

γ

, then relation between

α, β

and

γ

is-

Q. A vertical pole PS has two marks at Qand A such that the portions PQ, PR and PS subtend angles

α, β, γ

at a point on the ground distant x from the pole. If PQ = a, PR -b, PS = c and

α + β + γ

= 180; then

x^{2}

Join BYJU'S Learning Program

Let ∫x√1+x−x2 dx=α(1+x−x2)3/2+β[(x−12)√1+x−x2+γsin−12x−1√5]+C (where α,β,γ are constants). Then the value of 3(β−γα) is (where C is integration constant)

Let $\int x \sqrt{1 + x - x^{2}} d x = α (1 + x - x^{2})^{3 / 2} + β [(x - \frac{1}{2}) \sqrt{1 + x - x^{2}} + γ {sin}^{- 1} \frac{2 x - 1}{\sqrt{5}}] + C$ (where $α, β, γ$ are constants). Then the value of $3 (\frac{β - γ}{α})$ is
$($ where $C$ is integration constant $)$