What is the integral from-1 to 1 of cos^4(x) ?

The integral from-1 to 1 of cos^4(x) is 2cos^3(1)sin(1)+3/4-3/16 sin(4)

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integral from-1 to 1 of cos^4(x)

\int_{- 1}^{1} cos^{4} (x) d x

2 cos^{3} (1) sin (1) + \frac{3}{4} - \frac{3}{16} sin (4)

Decimal Notation

1.15734 \dots

Solution steps

\int_{- 1}^{1} cos^{4} (x) d x

Simplify

cos^{4} (x) : cos^{3} (x) cos (x)

= \int_{- 1}^{1} cos^{3} (x) cos (x) d x

Apply Integration By Parts

: [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{- 1}^{1}

= [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{- 1}^{1}

\int - 3 cos^{2} (x) sin^{2} (x) d x = - \frac{3}{8} (x - \frac{1}{4} sin (4 x))

= [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{- 1}^{1}

Simplify

= [cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))]_{- 1}^{1}

Compute the boundaries:

2 cos^{3} (1) sin (1) + \frac{3}{4} - \frac{3}{16} sin (4)

= 2 cos^{3} (1) sin (1) + \frac{3}{4} - \frac{3}{16} sin (4)

\int_{0}^{4} π (x)^{2} d x

\int_{0}^{x^{4}} e^{t} d t

\int_{3}^{4} x^{2} - 3 x d x

\int_{0}^{3} e^{- 3 x} d x

\int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

What is the integral from-1 to 1 of cos^4(x) ?
The integral from-1 to 1 of cos^4(x) is 2cos^3(1)sin(1)+3/4-3/16 sin(4)