Antiderivative involving Trig Identities.

[shamieh]

A little confused on something.

Suppose I have the integral

\(\displaystyle 2 \int 4 \sin^2x \, dx\)

So I understand that \(\displaystyle \sin^2x = \frac{1 - \cos2x}{2}\)

BUT we have a 4 in front of it, so shouldn't we pull the \(\displaystyle 4\) out in front of the integral to get:

\(\displaystyle 8 \int \frac{1 - \cos 2x}{2} \, dx\)

then pull out the \(\displaystyle \frac{1}{2}\) to get: \(\displaystyle 4 \int 1 - \cos 2x \, dx\)

then:

\(\displaystyle 8 [ x + \frac{1}{2} \sin2x ] + C\)?

 

[MarkFL]

I would switch the constants so that we have:

\(\displaystyle 4\int 2\sin^2(x)\,dx\)

Apply the identity:

\(\displaystyle 4\int 1-\cos(2x)\,dx\)

And then we have:

\(\displaystyle 4x-2\sin(2x)+C\)

 

[Prove It]

A little confused on something.

Suppose I have the integral

\(\displaystyle 2 \int 4 \sin^2x \, dx\)

So I understand that \(\displaystyle \sin^2x = \frac{1 - \cos2x}{2}\)

BUT we have a 4 in front of it, so shouldn't we pull the \(\displaystyle 4\) out in front of the integral to get:

\(\displaystyle 8 \int \frac{1 - \cos 2x}{2} \, dx\)

then pull out the \(\displaystyle \frac{1}{2}\) to get: \(\displaystyle 4 \int 1 - \cos 2x \, dx\)

then:

\(\displaystyle 8 [ x + \frac{1}{2} \sin2x ] + C\)?

Surely you mean $\displaystyle \begin{align*} 4 \left[ x + \frac{1}{2}\sin{(2x)} \right] + C \end{align*}$. Apart from that everything you have done is correct :)

 

[shamieh]

I would switch the constants so that we have:

\(\displaystyle 4\int 2\sin^2(x)\,dx\)

Apply the identity:

\(\displaystyle 4\int 1-\cos(2x)\,dx\)

And then we have:

\(\displaystyle 4x-2\sin(2x)+C\)

I see. Ok here is where I am getting lost Mark. My original problem was this \(\displaystyle \int \frac{x^2}{\sqrt{4 - x^2}} dx\)

So I got to the step we just mentioned above (ignore that my final answer had an 8 in front of all those terms, should be a 4 - BUT I then noticed that the A.D. of \(\displaystyle \cos(2\theta)\) is just \(\displaystyle \frac{1}{2}\sin(2\theta) \)THUS I can now change the 4 to a 2 since\(\displaystyle \frac{4}{2} = 2\) when I pulled it out to the constant.)

Finally, after applying all of those changes - now I am here:

\(\displaystyle 2\theta - 2\sin(2\theta) + C\) ... So then I said ok, simple enough, just sub back in for my original equation where I had \(\displaystyle x = 2\sin\theta\) which becomes \(\displaystyle \theta = \arcsin(\frac{x}{2})\) (Note: From when I used my trig sub)

Thus \(\displaystyle 2\arcsin(\frac{x}{2}) - 2\sin(2\arcsin(\frac{x}{2})) \) But somehow my teacher is getting:

\(\displaystyle 2\arcsin(\frac{x}{2}) - \frac{1}{2} x\sqrt{4 -x^2} + C\) as the final answer..In the second term when i have sin(arcsin ... etc What would I do there?

 

[MarkFL]

We are given:

\(\displaystyle I=\int \frac{x^2}{\sqrt{4-x^2}}\,dx\)

I would use the substitution:

\(\displaystyle x=2\sin(\theta)\,\therefore\,dx=2\cos(\theta)\,d \theta\)

And so now we have:

\(\displaystyle I=\int \frac{4\sin^2(\theta)}{\sqrt{4-4\sin^2(\theta)}}\,2\cos(\theta)\,d\theta\)

Simplifying this, we obtain:

\(\displaystyle I=4\int \sin^2(\theta)\,d\theta\)

This is half of what I began with in post #2 above, so using that method, we would obtain:

\(\displaystyle I=2\theta-\sin(2\theta)+C\)

Now, using the double-angle identity for sine, we may write:

\(\displaystyle I=2\theta-2\sin(\theta)\cos(\theta)+C\)

Observing that:

\(\displaystyle \sin(\theta)=\frac{x}{2}\implies\cos(\theta)=\frac{\sqrt{4-x^2}}{2}\)

we may now write:

\(\displaystyle I=2\sin^{-1}\left(\frac{x}{2} \right)-\frac{x\sqrt{4-x^2}}{2}+C\)

I think you are still trying to pull constants out in front of your integrals that are not factors of the entire integrand. :D

 

[shamieh]

Yes you are exactly right. I was pulling things out in front that I couldn't (Drunk)

So essentially I should have

\(\displaystyle 4[\theta - \frac{1}{2}\sin(2\theta)]\)

Which becomes:

\(\displaystyle 4\theta-2\sin(2\theta)\)

Which then becomes:

\(\displaystyle 2\theta - \sin(2\theta)\)

Then plug in double angle formula,

Then 2 in the second term cancels out with the \(\displaystyle \frac{x}{2}\)

Wow I see where I made my many errors. Thanks for the clarification.