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Integration involving Inverse Hyperbolic Functions

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Hi, could anybody help me with this integration question? Thing is, I understand the integration process through substitution x = 2sin(u) but there is a notice box in my book suggesting that this could be done without using the substitution x = 2sin(u). I have attached the question from my book as an image. I would be really thankful if anyone could help.

q4333.png
 
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PlanetMaster

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It is just referring to standard integrals that we can use for common integrals.

For example, you probably already know that
\(\int \frac{1}{x}\mathop{dx}=\ln x+C \)

Similarly, there is another standard integral
\(\int \frac{1}{\sqrt{1-x^{2}}}\mathop{dx}=\sin^{-1}{x+C} \)

We can compare the question with this standard integral and manipulate
\(\int{\sqrt{4-x^{2}}}\mathop{dx} \Leftrightarrow\int \frac{{4-x^{2}}}{\sqrt{4-x^{2}}}\mathop{dx} \)

Proceeding with integrating by parts using our standard integral (and not doing any actual integration this way), this gives
\(\frac{1}{2}\sqrt{{4-x^{2}}}x+{2sin^{-1}}\frac{x}{2}+C \)

I'm attaching a list of some of the standard integrals.
There are even more standard integrals out there that could solve the question in just 3 steps but depending on your paper and awarding body, you probably don't need to know these at this stage.
 

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Thing is, I tried doing it through integration by parts but it does not work. It only works if I use the substitution. Could you please share the steps if it can be done by integration by parts without substitution. I would be really grateful. PlanetMaster
 
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Here are my steps.......

Integration By Parts Formula I am using:
formula ibp.png

Step 1:
step 1.png

Step 2:
step 2.png

Step 3:
step 3.png

Again we will apply integration by parts.......

Then Step 4:
step 4.png

Step 5:
step 5.png

Should I continue further?........ PlanetMaster
 
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Step 6:
step 6.png

Again using integration by parts:
Step 7:
step 7.png

Step 8:
step 8.png

Now we get back the integral in step 3, I am a bit confused, any clue about what went wrong?..... PlanetMaster
 

PlanetMaster

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I'll try with steps but using the above standard integral will still require some sort of substitution eventually (albeit a much easier one).

Starting off with
\(\int \frac{{4-x^{2}}}{\sqrt{4-x^{2}}}\mathop{dx} \)

Integrating by parts, we have
\(u=4-x^{2} \)
\(du=-2x\mathop{dx} \)
\(dv=(4-x^{2})^{-\frac{1}{2}}\mathop{dx} \)
\(v=sin^{-1}(\frac{x}{2})+C \)

(This \(v \) above is what your book was referring to)

Substituting above in integration by parts equation, we get
\((4-x^{2})(sin^{-1}(\frac{x}{2}))-\int((sin^{-1}(\frac{x}{2}))(-2x\mathop{dx})) \)

Simplifying this, we get
\((4-x^{2})(sin^{-1}(\frac{x}{2}))+2\int x(sin^{-1}(\frac{x}{2}))\mathop{dx} \)

Now this is a fairly simple integral that we can solve either using u=x/2 substitution or by parts again (since we have only x now as u).
I'll skip the steps here but this should give (if needed, use any online integral compute tool to get steps)

\((4-x^{2})(sin^{-1}(\frac{x}{2}))+\frac{1}{2}x\sqrt{4-x^{2}}+(-2+x^{2})sin^{-1}(\frac{x}{2})+C \)

And this simplifies to
\(\frac{1}{2}\sqrt{{4-x^{2}}}x+{2sin^{-1}}\frac{x}{2}+C \)

Hope this helps!
 
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I tried to do it using the substitution u=x/2. It gets the required answer but through a trig substitution again which we don't want to use. Although, if I apply integration by parts it is working and leads to the desired result without substitution. I have attached my steps below. PlanetMaster
 
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Step 1:
step 1.png

Step 2:
step 2.png

Step 3:
step 3.png

Step 4:
step 4.png

Step 5:
step 5.png

Step 6:
step 6.png

Step 7:
step 7.png

Step 8:
step 8.png

Step 9:
step 9.png

Step 10:
step 10.png
 
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