Alternate method for integral of x^2/sqrt(5-x^2)

[knowledgeseeker]

Hello i was trying to solve the integral of (x^2)/(sqrt(5-4x^2 )) dx by ysing the substitution u=sqrt(5-4x^2) dx = (sqrt(5-4x^2)/-4x ) du So the integral can be written as Integral of -x/4 du and x= sqrt((5-u^2)/4) when x is >0 and x=-sqrt((5-u^2)/4) when x<0, i rewritten the integral as -1/8*Integral of sqrt(5-u^2)du using trig substitution with u= sqrt(5)*sin(t) we have -5/16 *t - 5/16 *sin(t)*cos(t) t=arcsin(u^2/sqrt(5)). if we now reverse the substitutions and simplify i got the final answer as -5/16 arcsin(sqrt((5-4x^2)/5)) - xsqrt(5-4x^2)/8 However this differs from the answer in my textbook which i will apload an image of(my textbook used different methods that is simpler and easier but i want to know where i made a mistake).

 

[blamocur]

i rewritten the integral as -1/8*Integral of sqrt(5-u^2)du

I don't understand how you got this part. In general, I find your post very difficult to read. At the very least, use separate lines for different steps. Also, I could not find an expression in your post for [imath]dx[/imath] in terms of [imath]u[/imath] and [imath]du[/imath].

 

[stapel]

Hello i was trying to solve the integral of (x^2)/(sqrt(5-4x^2 )) dx by ysing the substitution u=sqrt(5-4x^2) dx = (sqrt(5-4x^2)/-4x ) du So the integral can be written as Integral of -x/4 du and x= sqrt((5-u^2)/4) when x is >0 and x=-sqrt((5-u^2)/4) when x<0, i rewritten the integral as -1/8*Integral of sqrt(5-u^2)du using trig substitution with u= sqrt(5)*sin(t) we have -5/16 *t - 5/16 *sin(t)*cos(t) t=arcsin(u^2/sqrt(5)). if we now reverse the substitutions and simplify i got the final answer as -5/16 arcsin(sqrt((5-4x^2)/5)) - xsqrt(5-4x^2)/8 However this differs from the answer in my textbook which i will apload an image of(my textbook used different methods that is simpler and easier but i want to know where i made a mistake).View attachment 35089

Your formatting is a bit unclear to me. Is the following an accurate restatement of the above?

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Solve the following integral:

[imath]\qquad \int \frac{x^2}{\sqrt{5-4x^2}} dx[/imath]

I was trying to solve this integral by using the following substitution:

[imath]\qquad u = \sqrt{5-4x^2}\,dx = \frac{\sqrt{5-4x^2}}{-4x}\,du \qquad (1)[/imath]

So the integral can be written as:

[imath]\qquad \int \frac{-x}{4}\, du \qquad (2)[/imath]

...where:

[imath]\qquad x = \begin{cases} \sqrt{\frac{5-u^2}{4}} & \text{ for } x > 0 \\ -\sqrt{\frac{5-u^2}{4}} & \text{ for } x < 0 \end{cases} \qquad (3)[/imath]

I have rewritten the integral as:

[imath]\qquad -\frac{1}{8}\, \int \sqrt{5-u^2}\,du \qquad (4)[/imath]

...using trig substitution with:

[imath]\qquad u = \sqrt{5}\cdot \sin(t) \qquad (5)[/imath]

We have the following:

[imath]\qquad -\frac{5}{16}\cdot t - \frac{5}{16} \sin(t)\cos(t) \qquad (6)[/imath]

...where:

[imath]\qquad t = \arcsin\left(\frac{u^2}{\sqrt{5}}\right) \qquad (7)[/imath]

If we now reverse the substitutions and simplify, I got the final answer as:

[imath]\qquad -\frac{5}{16}\, \arcsin\left(\sqrt{\frac{5-4x^2}{5}}\right) - \frac{x\,\sqrt{5-4x^2}}{8} \qquad (8)[/imath]

However this differs from the answer in my textbook which i will apload an image of(my textbook used different methods that is simpler and easier but i want to know where i made a mistake).

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Is the above correct? Thank you!

Eliz.

 

[Dr.Peterson]

It turns out that your answer is equivalent to theirs.

That's because if [imath]\sin(\theta)=\sqrt{\frac{5-4x^2}{5}}[/imath], then [imath]\cos(\theta)=\frac{2x}{\sqrt{5}}[/imath], so [math]\arcsin\left(\sqrt{\frac{5-4x^2}{5}}\right)=\arccos\left(\frac{2x}{\sqrt{5}}\right)=\frac{\pi}{2}-\arcsin\left(\frac{2x}{\sqrt{5}}\right)[/math]
The constant absorbs the [imath]\frac{\pi}{2}[/imath], so this accounts for both the sign change and the argument change.

There is one typo in your work; in line (7) you put [imath]u^2[/imath] where you meant (and later used) [imath]u[/imath].

On the other hand, what happened to the sign issues you pointed out in line (3)?

 

[farhan]

One alternate method to solve the integral ∫(x^2/√(5-x^2)) dx is to use the trigonometric substitution x = √(5)sinθ.

Substituting x = √(5)sinθ, we have:

dx/dθ = √(5)cosθ

x^2 = 5sin^2θ

√(5-x^2) = √(5-5sin^2θ) = √(5)cosθ

Making these substitutions, the integral becomes:

∫(x^2/√(5-x^2)) dx = ∫[(5sin^2θ)/(√(5)cosθ)] (√(5)cosθ dθ/√(5)cosθ)

Simplifying, we get:

∫(5sin^2θ) dθ

Using the identity sin^2θ = (1-cos2θ)/2, we have:

∫(5sin^2θ) dθ = ∫[(5/2)(1-cos2θ)] dθ

= (5/2)(θ - (1/2)sin2θ) + C

Substituting back x = √(5)sinθ, we have:

∫(x^2/√(5-x^2)) dx = (5/2)arcsin(x/√5) - (x/2)√(5-x^2) + C

Therefore, the integral can also be expressed as (5/2)arcsin(x/√5) - (x/2)√(5-x^2) + C using the trigonometric substitution method.