# Integration involving trigonometric functions

#### bergausstein

##### Active member
any hints on how to work out this problems.

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

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

thanks!

#### MarkFL

Staff member
Re: Integration of trig

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...

#### bergausstein

##### Active member
Re: Integration of trig

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...
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.).

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

#### MarkFL

Staff member
Re: Integration of trig

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:

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

#### bergausstein

##### Active member
Re: Integration of trig

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.

#### MarkFL

Staff member
Re: Integration of trig

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.
Well, I showed you how I would approach it.

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.

#### MarkFL

Staff member
Re: Integration of trig

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.

#### Prove It

##### Well-known member
MHB Math Helper
Re: Integration of trig

any hints on how to work out this problems.

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

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

thanks!
\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*}

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

#### Prove It

##### Well-known member
MHB Math Helper
Re: Integration of trig

any hints on how to work out this problems.

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

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

thanks!
\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*}

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