MHB Antiderivative involving Trig Identities.

[shamieh]

A little confused on something.

Suppose I have the integral

$$2 \int 4 \sin^2x \, dx$$

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

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

$$8 \int \frac{1 - \cos 2x}{2} \, dx$$

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

then:

$$8 [ x + \frac{1}{2} \sin2x ] + C$$?

 

[MarkFL]

I would switch the constants so that we have:

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

Apply the identity:

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

And then we have:

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

 

[Prove It]

A little confused on something.

Suppose I have the integral

$$2 \int 4 \sin^2x \, dx$$

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

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

$$8 \int \frac{1 - \cos 2x}{2} \, dx$$

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

then:

$$8 [ x + \frac{1}{2} \sin2x ] + C$$?

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:

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

Apply the identity:

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

And then we have:

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

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

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 $$\cos(2\theta)$$ is just $$\frac{1}{2}\sin(2\theta) $$THUS I can now change the 4 to a 2 since$$ \frac{4}{2} = 2$$ when I pulled it out to the constant.)

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

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

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

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

 

[MarkFL]

We are given:

$$I=\int \frac{x^2}{\sqrt{4-x^2}}\,dx$$

I would use the substitution:

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

And so now we have:

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

Simplifying this, we obtain:

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

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

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

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

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

Observing that:

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

we may now write:

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

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

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

Which becomes:

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

Which then becomes:

$$2\theta - \sin(2\theta)$$

Then plug in double angle formula,

Then 2 in the second term cancels out with the $$\frac{x}{2}$$

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