"Evaluate integral : ∫cos4(x)dx" was answered by our experts

Concepcion Hale

Answered question

2022-01-03

Evaluate integral :

\int \cos^{4} (x) d x

Answer & Explanation

Bertha Jordan

Beginner2022-01-04Added 37 answers

\cos^{4} (x) = {(\frac{1 + \cos (2 x)}{2})}^{2} = \frac{1 + \cos^{2} (2 x) + 2 \cos (2 x)}{4}

= \frac{1 + \frac{1 + \cos (4 x)}{2} + 2 \cos (2 x)}{4}

which gives us

\cos^{4} (x) = \frac{3 + 4 \cos (2 x) + \cos (4 x)}{8}

Now you should be able to integrate this off.

jean2098

Beginner2022-01-05Added 38 answers

Using the reduction formulae,

\int \cos^{n} x d x = \frac{\cos^{n - 1} x \sin x}{n} + \frac{n - 1}{n} \int \cos^{n - 2} x d x

Putting

n = 2

\int \cos^{2} x d x = \frac{\cos x \sin x}{2} + \frac{1}{2} \int d x = \frac{\cos x \sin x}{2} + \frac{1}{2} x + C

Putting

n = 4

\int \cos^{4} x d x = \frac{\cos^{3} x \sin x}{4} + \frac{3}{4} \int \cos^{2} x d x

Vasquez

Expert2022-01-09Added 669 answers

$\cos^{4} x = \cos^{2} x - \cos^{2} x \sin^{2} x \Rightarrow$
$\Rightarrow I := \int \cos^{4} x d x = \int \cos^{2} x d x - \int \cos^{2} x \sin^{2} x d x$
$= \frac{x + \cos x \sin x}{2} + \int \sin x \cos^{2} x (- \cos)^{'} x d x$
Now by parts in the last integegral:
$u = \sin x, u^{'} = \cos x$
$v^{'} = \cos^{2} x \sin x, v = - \frac{1}{3} \cos^{3} x$
so
$I := \frac{x \cos x \sin x}{2} - \frac{1}{3} \cos^{3} x \sin x + \frac{1}{3} I \Rightarrow . . .$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?