What is the integral from 0 to pi/2 of 4cos^4(x) ?

The integral from 0 to pi/2 of 4cos^4(x) is (3pi)/4

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integral from 0 to pi/2 of 4cos^4(x)

\int_{0}^{\frac{π}{2}} 4 cos^{4} (x) d x

\frac{3 π}{4}

Decimal

2.35619 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 4 cos^{4} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{4} (x) d x

Simplify

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

= 4 \cdot \int_{0}^{\frac{π}{2}} 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]_{0}^{\frac{π}{2}}

= 4 [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{0}^{\frac{π}{2}}

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

= 4 [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{\frac{π}{2}}

Simplify

4 [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{\frac{π}{2}} : 4 [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{2}}

= 4 [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{3 π}{16}

= 4 \cdot \frac{3 π}{16}

Simplify

= \frac{3 π}{4}

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What is the integral from 0 to pi/2 of 4cos^4(x) ?
The integral from 0 to pi/2 of 4cos^4(x) is (3pi)/4