MHB Can I Simplify Trigonometric Integrals by Taking out Constants?

[shamieh]

Quick question.

$$\int sin^{4}x dx$$

so I know:

$$\frac{1}{2} \int 1 - 2cos2x + \frac{1}{2}(1 + cos4x)dx$$

So here I first brought out the 1/2 because it's a constant and it's nasty.

so now I have

$$\frac{1}{4} \int 1 - 2cos2x + 1 + cos4x dx$$

so...Just as I brought 1/2 out can I now precede to take the 2 constant out that is in front of 2cos2x? thus turning it into:

$$\frac{1}{2} \int 1 - cos2x + 1 + cos4x dx$$ ?

- - - Updated - - -

Ah wait, I think I forgot to square the denominator of (1 - cos2x/2)^2 because it was sin^4

 

[MarkFL]

A factor brought out in front of the integral must be a factor of the entire integrand. For example:

$$\sin^4(x)=\frac{1}{8}\left(3-4\cos(2x)+\cos(4x) \right)$$

And so we may write:

$$\int \sin^4(x)\,dx=\frac{1}{8}\int 3-4\cos(2x)+\cos(4x)\,dx$$

 

[shamieh]

A factor brought out in front of the integral must be a factor of the entire integrand. For example:

$$\sin^4(x)=\frac{1}{8}\left(3-4\cos(2x)+\cos(4x) \right)$$

And so we may write:

$$\int \sin^4(x)\,dx=\frac{1}{8}\int 3-4\cos(2x)+\cos(4x)\,dx$$

OH I see, so you can't bring the 3 out because it's not a factor.

 

[shamieh]

$$\frac{1}{4} \int 1 - 2cos2x + \frac{1}{2}(1 + cos4x) \ dx$$

So now I have

$$\frac{1}{4} \int \frac{3}{2} - 2cos2x + \frac{1}{2} + \frac{1}{2}cos4x) \ dx$$

which is $$\frac{1}{4} \int \frac{3}{2} - 2cos2x + \frac{1}{2}cos4x \ dx$$Figured out what to do. Thanks for the help. Integrated by parts the 2cos2x and the 1/2cos4x which were extremely similar so basically did it in my head to get the 1/8.

 

[MarkFL]

You really don't need integration by parts to integrate something of the form:

$$\int \cos(ax)\,dx$$ where $a$ is a non-zero real constant. Observing that:

$$\frac{d}{dx}\left(\frac{1}{a}\sin(ax) \right)=\cos(ax)$$

We may then simply write:

$$\int \cos(ax)\,dx=\frac{1}{a}\sin(ax)+C$$