Question

what is the formula of cos⁴x

Answer 1

Answer:

Step-by-step explanation:

Elegant problem.

Note: The reduction formulae.

[math]\boxed{\displaystyle\int \sin^{n} x \, dx = -\dfrac{1}{n} \sin^{(n-1)}x \cos x + \dfrac{n-1}{n} \int \sin^{n-2} x \, dx} \tag*{(1)}[/math]

[math]\boxed{\displaystyle \int \cos^{n} x \, dx = \dfrac{1}{n}\cos^{(n-1)}x \sin x + \dfrac{n-1}{n} \int \cos^{n-2} x \, dx} \tag*{(2)}[/math]

Now, the given integral is,

[math]\displaystyle I= \int_{0}^{2 \pi} \cos^{4}x \, dx \tag*{}[/math]

We take the indefinite integral as,

[math]\displaystyle I_{1}= \int \cos^{4}x \, dx \tag*{}[/math]

This can be written as, [for [math]n = 4[/math]]

[math]\displaystyle I_{1} = \int \cos^{4} x \, dx \\ = \dfrac{1}{4}\cos^{(4-1)}x \sin x + \dfrac{4-1}{4} \int \cos^{4-2} x \, dx \tag*{}[/math]

[math]\boxed{\displaystyle \implies I_{1} = \dfrac{1}{4}\cos^{3}x \sin x + \dfrac{3}{4} \int \cos^{2} x \, dx} \tag*{}[/math]

Now, take [math]I_{2}[/math] as,

[math]\displaystyle I_{2} = \int \cos^{2} x \, dx \tag*{}[/math]

[math]\displaystyle =\int \cos^{2} x \, dx \tag*{}[/math]

[math]\displaystyle =\int \dfrac{\cos 2x +1}{2} \, dx \tag*{}[/math]

Solve [math]I_{2}[/math]; plug it into [math]I_{1}[/math] and then substitute the limits.

In case you face any difficulty; please let me know.

Cheers!

Answer 2

cos⁴x=(cos²x)²

=(1-sin²x)²

=(1)²-2(1)(sin²x)+(sin²x)²