dx-9-4x-2-using-the-trigonometric-substitution-

Question Number 171910 by Tawa11 last updated on 21/Jun/22

\int \frac{dx}{9 - {4 x}^{2}}

using the trigonometric substitution .

Answered by aleks041103 last updated on 21/Jun/22

\int \frac{d x}{9 - 4 x^{2}} = \frac{1}{9} \int \frac{d x}{1 - {(\frac{2 x}{3})}^{2}} = \frac{1}{9} \frac{3}{2} \int \frac{d u}{1 - u^{2}} =

= \frac{1}{6} \int \frac{d u}{1 - u^{2}}

\frac{1}{1 - u^{2}} = \frac{1}{(1 - u) (1 + u)} = \frac{1}{2} (\frac{1}{1 - u} + \frac{1}{1 + u})

\int \frac{d u}{1 - u^{2}} = \frac{1}{2} l n ∣ 1 + u ∣ - \frac{1}{2} l n ∣ 1 - u ∣

\Rightarrow \int \frac{d x}{9 - 4 x^{2}} = \frac{1}{12} l n ∣ \frac{1 + \frac{2}{3} x}{1 - \frac{2}{3} x} ∣

\Rightarrow \int \frac{d x}{9 - 4 x^{2}} = \frac{1}{12} l n ∣ \frac{3 + 2 x}{3 - 2 x} ∣

Commented by Tawa11 last updated on 21/Jun/22

I appreciate sir .

Commented by Tawa11 last updated on 21/Jun/22

But sir, I need the trigonometric [substion .

like . let x = 3 \tan θ

Commented by Tawa11 last updated on 21/Jun/22

Can the question be solved using trigonometric substitution ?

Commented by mr W last updated on 21/Jun/22

c e r t a i n l y y o u c a n! t h a t m e a n s y o u

c a n a l w a y s m a k e e a s y t h i n g s m o r e

c o m p l i c a t e d, w h e n y o u l i k e .

Commented by Tawa11 last updated on 21/Jun/22

Because the condition given is using trigonometric substitution sir .

Commented by mr W last updated on 21/Jun/22

y o u c a n d o i t! i j u s t s a i d i t i s n o t a

g o o d i d e a . t r y w i t h u = \frac{3}{2} \tan θ, y o u

w i l l s e e .

Commented by Tawa11 last updated on 21/Jun/22

Thanks sir .

Commented by Tawa11 last updated on 22/Jun/22

\int \frac{dx}{9 - {4 x}^{2}}

= \int \frac{\frac{3}{2} \sec^{2} θ}{9 - 9 \tan^{2} θ} d θ

= \frac{3}{18} \int \frac{\sec^{2} θ}{1 - \tan^{2} θ} d θ

\sec^{2} θ = 1 + \tan^{2} θ

From here, please what next \dots

Answered by ajfour last updated on 22/Jun/22

l e t I = \int \frac{d x}{9 - 4 x^{2}} = 4 \int \frac{d x}{36 - 16 x^{2}}

= 4 \int \frac{d x}{(6 - 4 x) (6 + 4 x)}

= \frac{4}{12} \int \frac{(6 - 4 x) + (6 + 4 x)}{(6 - 4 x) (6 + 4 x)} d x

= \frac{1}{3} \int \frac{d x}{6 + 4 x} + \frac{1}{3} \int \frac{d x}{6 - 4 x}

= \frac{1}{12} \int \frac{4 d x}{6 + 4 x} - \frac{1}{12} \int \frac{- 4 d x}{6 - 4 x}

= \frac{1}{12} \ln ∣ 6 + 4 x ∣ - \frac{1}{12} \ln ∣ 6 - 4 x ∣ + c

= \frac{1}{12} \ln ∣ \frac{3 + 2 x}{3 - 2 x} ∣ + c

Commented by Tawa11 last updated on 22/Jun/22

God bless you sir .

Answered by thfchristopher last updated on 22/Jun/22

Let x = \frac{3}{2} \sin θ

d x = \frac{3}{2} \cos θ d θ

\sin θ = \frac{2 x}{3}, \tan θ = \frac{2 x}{\sqrt{9 - 4 x^{2}}}, \sec θ = \frac{3}{\sqrt{9 - 4 x^{2}}}

∴ \int \frac{d x}{9 - 4 x^{2}}

= \frac{3}{2} \int \frac{\cos θ}{{9 \cos}^{2} θ} d θ

= \frac{1}{6} \int \sec θ d θ

= \frac{1}{6} \ln ∣ \tan θ + \sec θ ∣ + C

= \frac{1}{6} \ln ∣ \frac{3 + 2 x}{\sqrt{9 - 4 x^{2}}} ∣ + C

= \frac{1}{6} \ln ∣ \sqrt{\frac{3 + 2 x}{3 - 2 x}} ∣ + C

= \frac{1}{12} \ln ∣ \frac{3 + 2 x}{3 - 2 x} ∣ + C

Commented by thfchristopher last updated on 22/Jun/22

Trigonometric substitution applied on

integral format :

(i) \int \frac{d x}{a^{2} + b^{2} x^{2}} or \int \frac{d x}{b^{2} x^{2} + a^{2}}

Let b x = a \tan θ

(ii) \int \frac{d x}{a^{2} - b^{2} x^{2}}

Let b x = a \sin θ

(iii) \int \frac{d x}{b^{2} x^{2} - a^{2}}

Let b x = a \sec θ

Commented by Tawa11 last updated on 22/Jun/22

God bless you sir

Commented by Tawa11 last updated on 22/Jun/22

I really appreciate .

Commented by mr W last updated on 22/Jun/22

a s u b s t i t u t i o n {d o e s n}^{'} t m a k e t h e t h i n g

e a s i e r i n a n y f o r m, t h e r e f o r e u n n e c e s a r y .

e . g .

\int \frac{d x}{a^{2} - b^{2} x^{2}}

= \frac{1}{2 a} \int (\frac{1}{a - b x} + \frac{1}{a + b x}) d x

= \frac{1}{2 a b} [- \ln (a - b x) + \ln (a + b x)]

= \frac{1}{2 a b} \times \ln ∣ \frac{a + b x}{a - b x} ∣ + C

\int \frac{d x}{b^{2} x^{2} - a^{2}} = - \int \frac{d x}{a^{2} - b^{2} x^{2}} = - \frac{1}{2 a b} \times \ln ∣ \frac{a + b x}{a - b x} ∣ + C

Commented by Tawa11 last updated on 22/Jun/22

God bless you sir . I really appreciate .

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