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#### bergausstein

##### Active member

- Jul 30, 2013

- 191

$\displaystyle\int\frac{dx}{(1-sinx)^2}$

$\displaystyle\int\sin x\sin2x\sin3x dx$

thanks!

- Thread starter bergausstein
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- Thread starter
- #1

- Jul 30, 2013

- 191

$\displaystyle\int\frac{dx}{(1-sinx)^2}$

$\displaystyle\int\sin x\sin2x\sin3x dx$

thanks!

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1.) I would try the following:

Use a Pythagorean identity and the double-angle identity for sine:

\(\displaystyle 1-\sin(x)=\sin^2\left(\frac{x}{2} \right)-2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)+\cos^2\left(\frac{x}{2} \right)\)

Factor as the square of a binomial:

\(\displaystyle 1-\sin(x)=\left(\sin\left(\frac{x}{2} \right)-\cos\left(\frac{x}{2} \right) \right)\)

Apply a linear combination identity:

\(\displaystyle 1-\sin(x)=2\sin^2\left(\frac{x}{2}-\frac{\pi}{4} \right)\)

Now the integral is:

\(\displaystyle I=\frac{1}{4}\int \csc^4\left(\frac{x}{2}-\frac{\pi}{4} \right)\,dx\)

Use the substitution:

\(\displaystyle u=\frac{x}{2}-\frac{\pi}{4}\,\therefore\,du=\frac{1}{2}dx\)

And we have:

\(\displaystyle I=\frac{1}{2}\int \csc^4(u)\,du\)

Using the Pythagorean identity:

\(\displaystyle \csc^2(\theta)=1+\cot^2(\theta)\) we may write:

\(\displaystyle I=\frac{1}{2}\int \left(1+\cot^2(u) \right)\csc^2(u)\,du\)

Now, you should see a good substitution to use to finish...

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- #3

- Jul 30, 2013

- 191

sure i can solve the new form of the problem. but the preceding steps are complicated. (e.g the linear combination identity, i haven't heard of it.).1.) I would try the following:

Use a Pythagorean identity and the double-angle identity for sine:

\(\displaystyle 1-\sin(x)=\sin^2\left(\frac{x}{2} \right)-2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)+\cos^2\left(\frac{x}{2} \right)\)

Factor as the square of a binomial:

\(\displaystyle 1-\sin(x)=\left(\sin\left(\frac{x}{2} \right)-\cos\left(\frac{x}{2} \right) \right)\)

Apply a linear combination identity:

\(\displaystyle 1-\sin(x)=2\sin^2\left(\frac{x}{2}-\frac{\pi}{4} \right)\)

Now the integral is:

\(\displaystyle I=\frac{1}{4}\int \csc^4\left(\frac{x}{2}-\frac{\pi}{4} \right)\,dx\)

Use the substitution:

\(\displaystyle u=\frac{x}{2}-\frac{\pi}{4}\,\therefore\,du=\frac{1}{2}dx\)

And we have:

\(\displaystyle I=\frac{1}{2}\int \csc^4(u)\,du\)

Using the Pythagorean identity:

\(\displaystyle \csc^2(\theta)=1+\cot^2(\theta)\) we may write:

\(\displaystyle I=\frac{1}{2}\int \left(1+\cot^2(u) \right)\csc^2(u)\,du\)

Now, you should see a good substitution to use to finish...

i believe there's an easier method to solve this.

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- #4

The linear combination identity is a good tool to have when dealing with expressions of the form:

\(\displaystyle a\sin(\theta)+b\cos(\theta)\)

It allows us the write the above sinusoidal expression as a constant times a single sinusoid. When $0<a$, we have:

\(\displaystyle \sqrt{a^2+b^2}\sin\left(\theta+\tan^{-1}\left(\frac{b}{a} \right) \right)\)

If $a<0$ then we have:

\(\displaystyle \sqrt{a^2+b^2}\sin\left(\theta+\tan^{-1}\left(\frac{b}{a} \right)+\pi \right)\)

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- #5

- Jul 30, 2013

- 191

yes, that's interesting. but still i want a more comprehensive method.

by the way i think it's good to say that I'm just beginning to learn this subject.

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- #6

Well, I showed you how I would approach it.yes, that's interesting. but still i want a more comprehensive method.

by the way i think it's good to say that I'm just beginning to learn this subject.

Perhaps someone else can show you a method which suits your tastes.

You will find that linear combination identity cropping up in many places, so I would sincerely advise you to incorporate it into those things you are learning.

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- #7

2.) Consider:

\(\displaystyle 4\sin(x)\sin(2x)\sin(3x)=-2\sin(3x)\left(-2\sin(2x)\sin(x) \right)\)

Applying a product to sum identity (these are also very useful), we may write:

\(\displaystyle 4\sin(x)\sin(2x)\sin(3x)=-2\sin(3x)\left(\cos(3x)-\cos(x) \right)\)

Distribute:

\(\displaystyle 4\sin(x)\sin(2x)\sin(3x)=2\sin(3x)\cos(x)-2\sin(3x)\cos(3x)\)

For the first term, apply another product to sum identity, and for the second term, the double-angle identity for sine:

\(\displaystyle 4\sin(x)\sin(2x)\sin(3x)=\sin(2x)+\sin(4x)-\sin(6x)\)

Hence:

\(\displaystyle \sin(x)\sin(2x)\sin(3x)=\frac{1}{4}\left(\sin(2x)+\sin(4x)-\sin(6x) \right)\)

Now integrate term by term.

[tex]\displaystyle \begin{align*} \int{ \frac{dx}{ \left[ 1 - \sin{(x)} \right] ^2 } } &= \int{ \frac{\left[ 1 + \sin{(x)} \right] ^2 \, dx }{ \left[ 1 - \sin{(x)} \right] ^2 \left[ 1 + \sin{(x)} \right] ^2 } } \\ &= \int{ \frac{ \left[ 1 + 2\sin{(x)} + \sin^2{(x)} \right] \, dx}{ \left[ 1 - \sin^2{(x)} \right] ^2 } } \\ &= \int{ \frac{ \left[ 1 + 2\sin{(x)} + \sin^2{(x)} \right] \, dx}{ \left[ \cos^2{(x)} \right] ^2 } } \\ &= \int{ \sec^4{(x)}\,dx } + 2\int{ \frac{\sin{(x)}\, dx}{\cos^4{(x)}} } + \int{ \tan^2{(x)}\sec^2{(x)} \, dx} \end{align*}[/tex]

$\displaystyle\int\frac{dx}{(1-sinx)^2}$

$\displaystyle\int\sin x\sin2x\sin3x dx$

thanks!

The first integral can be solved using a reduction formula, the second with the substitution [tex]\displaystyle \begin{align*} u = \cos{(x)} \end{align*}[/tex] and the third with the substitution [tex]\displaystyle \begin{align*} v = \tan{(x)} \end{align*}[/tex].

[tex]\displaystyle \begin{align*} \int{ \sin{(x)}\sin{(2x)}\sin{(3x)}\,dx} &= \int{ \sin{(x)} \left[ 2\sin{(x)}\cos{(x)} \right] \left[ -4\sin^3{(x)} + 3\sin{(x)} \right] \, dx} \end{align*}[/tex]

$\displaystyle\int\frac{dx}{(1-sinx)^2}$

$\displaystyle\int\sin x\sin2x\sin3x dx$

thanks!

Now this integral can be solved using the substitution [tex]\displaystyle \begin{align*} u = \sin{(x)} \end{align*}[/tex].