Trig Integral or Trig Substitution for This Problem?

[Ayesh]

Homework Statement

When I look at the equation, I see a trig integral.

\intsinxcos³x/\sqrt{1+(sin^2)x} dx

But the \sqrt{1++(sin^2)x} gets me confused.
I could tranform it into cos²x if it was sin²x - 1

The Attempt at a Solution

\intsinxcos³x/\sqrt{1++(sin^2)x} dx
=\intsintxcos²xcosx/\sqrt{1++(sin^2)x}
=\intsinx - sin²x/\sqrt{1++(sin^2)x}
=...?

 

[Mark44]

Homework Statement

When I look at the equation, I see a trig integral.

\intsinxcos³x/\sqrt{1+sin²x} dx

But the \sqrt{1+sin²x} gets me confused.
I could tranform it into cos²x if it was sin²x - 1

The Attempt at a Solution

\intsinxcos³x/\sqrt{1+sin²x} dx
=\intsintxcos²xcosx/\sqrt{1+sin²x}
=\intsinx - sin²x/\sqrt{1+sin²x}
=...?

This is really hard to read. You shouldn't mix [ sup] tags in between [ tex] tags. Instead, use code like ^2 or ^{2} for exponents within [ tex] tags.

 

[tt2348]

try a using the trig identity \sin (x)= \frac{1-\cos 2x}{2} and \cos x =\frac{1+\cos 2x}{2}
and
\sin x \cos x = 1/2 \sin 2x

it'll look like \int\frac{1}{2} \sin 2x *\frac{(1+\cos 2x)^{2}}{4}\frac{1}{\sqrt{1+1/4*(1+\cos 2x)^2}}
there's is a u sub that makes this a whole lot easier

 

[tt2348]

hint u=(1+cos2x) , du=-2sin2x

 

[Ayesh]

Thank you!

 

[Gib Z]

Perhaps even quicker is to use u=1 + \sin^2 x from the start.

 

[Ratio Test =)]

Recall that :

\sqrt{1+sin^2(x)}=\sqrt{1+1-cos^2(x)}=\sqrt{2-cos^2(x)}

Use t=cos(x).

Then do a trigonometric substitution.

 

[Ayesh]

Thank you!

 

[Ayesh]

Now that I am trying your hints, I feel even more lost...
Sorry.

I cannot use trig identities since my powers are odd (this is what is written in my notebook).

Well, I'll show you what I have done until now and then tell me if I am on the right way or not.

\int{sinxcos^3x/sqrt{1+sin^2x}

u=sinx
du=cosxdx

=\int{sinxcos^2xcosx/sqrt{1+sin^2x}
=\int{sinx(1-sin^2x)/sqrt{1+sin^2x}
==\int{u-u^2/1+u}
=...?

 

[Gib Z]

None of us even told you to do that substitution. If you actually carried out my suggestion you would have arrived at an elementary integral immediately.

 

[Ayesh]

Here is my new try:

\intsinxcos³x/\sqrt{1+sin^2(x)} dx

u=1+sin²x
du=1/2 cos²x

\intsinxcos²xcosx/\sqrt{1+sin^2(x)}

\int1/2 sinxcosx/\sqrt{1+sin^2(x)}

\int1/4 sin2x/\sqrt{1+sin^2(x)}

\int1/4 2u/\sqrt{u^1/2}

\int1/2 u^1/2

1/3 u^3/2

1/3(1+sin²x)^3/2


I know something isn't right, but I can't figure out why.

I know I'm not right since when I differentiate it my answer is sqrt(1+sin(t)^2)*sin(t)*cos(t)

 

[tt2348]

youre du is wrong.
derivative of sin^2(x)=2*sinxcosx

 

[tt2348]

i'm going to show you my general method for doing odd powered sinxcosx integrals, it might help you in other problems.
for some odd numbered natural numbers,k,and n

\int \sin^{k}(x)\cos^{n}(x) dx
since \sin x \cos x = \frac{1}{2}\sin (2x)
and \sin^{2} x = \frac{1-\cos 2x}{2}
and \cos^{2} x = \frac{1+\cos 2x}{2}
then \int \sin^{k}(x)\cos^{n}(x) dx =\int \sin^{k-1}(x)\cos^{n-1}(x)*(\sin x \cos x) dx <br /> =\int \left(\frac{1-\cos 2x}{2})\right)^{\frac{k-1}{2}}*\left(\frac {1+\cos 2x}{2}\right)^{\frac{n-1}{2}} *(\frac{1}{2}\sin (2x))dx
pick u=\frac{1-\cos 2x}{2} , du=2\sin 2x dx
then \frac{1}{2^{ \frac{k+n}{2}+1 } } \int u^{\frac{k-1}{2}} * \left(2-u\right)^{\frac{n-1}{2}} du

 

[Ayesh]

Even though I corrected my du, I don't feel like I'm getting closer the answer...
Thank you for trying to help the slow me!

 

[Bohrok]

What do you have now that you've corrected the du?

 

[vela]

It would really help if you provided more detail on what exactly you did and fixed typos and errors in what you write before posting it. We can't read your mind, so we can only go by what appears here. When there are numerous typos, it's difficult to figure out what you actually meant and what was just a careless mistake. Frankly, it's kind of annoying to have to try figure out what you did or what you mean because you're not willing to put the effort into expressing yourself clearly.

For example, you wrote

\intcos²x/\2sqrt{1+sin^2x}

Obviously, there's an error in the LaTeX code you wrote as the radical sign didn't appear correctly. What's less obvious is that you had a factor of 2 in front of the square root that didn't show up because of that error. The only reason I found this out was because the LaTeX appeared when I happened to hover the pointer over that part of the post.

\int1-sin²x/2\sqrt{1+sin^2x}

In this line, you fixed the LaTeX, and the two now appears. From our standpoint, the two just appeared mysteriously, and we wonder what exactly you did to get that two. So we can guess you did a substitution, replacing the dx by du which resulted in the two, but how are we supposed to know that? You didn't say what substitution you used, and you omitted any mention of dx and du in the integral. It could have just as easily been an algebra error, a mistaken trig identity, etc.

Finally, you need to follow the normal conventions of mathematics, particularly the use of parentheses. Without having seen where your second line came from, some would interpret your second line to mean

1-\frac{\sin^2 x}{2}\sqrt{1+\sin^2 x}

whereas you really (I think) meant

\frac{1-\sin^2 x}{2\sqrt{1+\sin^2 x}}

Writing (1-sin² x)/(2 sqrt(1+sin² x)) would have accurately conveyed your intent.

I hope you realize I'm not posting here just to criticize you or give you a hard time. In one sense, I just want you to be able to use this forum effectively. In a broader sense, I'm hoping you will understand how essential it is to be able to express your ideas accurately and clearly.