1-Evaluate-I-25x-2-4-x-dx-2-Find-value-of-2-5-4-5-f-x-dx-3-Find-value-of-4-5-2-5-f-x-dx-

Question Number 179383 by Acem last updated on 29/Oct/22

1 ∙ E v a l u a t e I = \int \frac{\sqrt{25 x^{2} - 4}}{x} d x

2 ∙ F i n d v a l u e o f \int_{\frac{2}{5}}^{\frac{4}{5}} f (x) d x

3 ∙ F i n d v a l u e o f \int_{- \frac{4}{5}}^{- \frac{2}{5}} f (x) d x

Answered by cortano1 last updated on 29/Oct/22

Y = \int \frac{\sqrt{{(5 x)}^{2} - 2^{2}}}{x} dx

let 5 x = 2 \sec t \Rightarrow dx = \frac{2}{5} \sec t \tan t dt

\sec t = \frac{5 x}{2} or \cos t = \frac{2}{5 x}

Y = \int \frac{\sqrt{4 \sec^{2} t - 4}}{(\frac{2}{5} \sec t)} . (\frac{2}{5} \sec t \tan t dt)

Y = \int {2 \tan}^{2} t dt

Y = 2 \int (\sec^{2} t - 1) dt

Y = 2 \tan t - 2 t + c

Y = 2 \frac{\sqrt{{25 x}^{2} - 4}}{2} - 2 \arccos (\frac{2}{5 x}) + c

Y = \sqrt{{25 x}^{2} - 4} - 2 \arccos (\frac{2}{5 x}) + c

or Y = \sqrt{{25 x}^{2} - 4} - 2 arcsec (\frac{5 x}{2}) + c

Commented by Acem last updated on 29/Oct/22

E x c e l l e n t!

Answered by Acem last updated on 29/Oct/22

2 ∙ \int_{\frac{2}{5}}^{\frac{4}{5}} \frac{\sqrt{25 x^{2} - 4}}{x} d x \dots (1)

\sqrt{25 x^{2} - 4} = 2 \sqrt{{(\frac{5 x}{2})}^{2} - 1}

\sec θ = \frac{5 x}{2} \Leftrightarrow x = \frac{2}{5} \sec θ \Rightarrow d x = \frac{2}{5} \sec θ \tan θ d θ

\frac{d x}{x} = \tan θ \dots (2)

2 \sqrt{{(\frac{5 x}{2})}^{2} - 1} = 2 \sqrt{\sec^{2} θ - 1} = 2 ∣ \tan θ ∣^{B e a w a k e!} \dots (3)

x = \frac{2}{5} \Rightarrow \sec θ = 1 \Rightarrow θ_{1} = 0

x = \frac{4}{5} \Rightarrow \sec θ = 2 \Rightarrow θ_{2} = \frac{π}{3}

a s 0 < θ < \frac{π}{3} \Rightarrow ∣ \tan θ ∣= \tan θ

b y c o m p e n s a t e 2, 3 i n 1 :

a = 2 \int_{0}^{\frac{π}{3}} (\sec^{2} θ - 1) d θ = 2 (\tan θ - θ) ∣_{0}^{\frac{π}{3}}

a = 2 (\sqrt{3} - \frac{π}{3})

3 ∙ N e x t

Answered by Acem last updated on 29/Oct/22

3 ∙ \int_{\frac{- 4}{5}}^{\frac{- 2}{5}} \frac{\sqrt{25 x^{2} - 4}}{x} d x \dots (1)

\sec θ = \frac{5 x}{2} \Leftrightarrow x = \frac{2}{5} \sec θ \Rightarrow d x = \frac{2}{5} \sec θ \tan θ d θ

\frac{d x}{x} = \tan θ \dots (2)

2 \sqrt{{(\frac{5 x}{2})}^{2} - 1} = 2 \sqrt{\sec^{2} θ - 1} = 2 ∣ \tan θ ∣^{B e a w a k e!} \dots (3)

x = \frac{- 4}{5} \Rightarrow \sec θ = - 2 \Rightarrow θ_{1} = \frac{2 π}{3}

x = \frac{- 2}{5} \Rightarrow \sec θ = - 1 \Rightarrow θ_{2} = π

a s \frac{2 π}{3} < θ < π \Rightarrow ∣ \tan θ ∣= - \tan θ

b y c o m p e n s a t e 2, 3 i n 1 :

a = - 2 \int_{\frac{2 π}{3}}^{π} (\sec^{2} θ - 1) d θ = - 2 (\tan θ - θ) ∣_{\frac{2 π}{3}}^{π}

a = 2 (\frac{π}{3} - \sqrt{3})

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