What is the integral from-pi/2 to pi/2 of 4cos^6(x) ?

The integral from-pi/2 to pi/2 of 4cos^6(x) is (5pi)/4

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integral from-pi/2 to pi/2 of 4cos^6(x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos^{6} (x) d x

\frac{5 π}{4}

Decimal

3.92699 \dots

Solution steps

\int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos^{6} (x) d x

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{6} (x) d x

Apply Integral Reduction

= 4 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{4} (x) d x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{4} (x) d x = \frac{3 π}{8}

= 4 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{8})

Simplify

4 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{8}) : 4 ([\frac{1}{6} cos^{5} (x) sin (x)]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{5 π}{16})

= 4 ([\frac{1}{6} cos^{5} (x) sin (x)]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{5 π}{16})

Compute the boundaries:

0

= 4 (0 + \frac{5 π}{16})

Simplify

= \frac{5 π}{4}

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What is the integral from-pi/2 to pi/2 of 4cos^6(x) ?
The integral from-pi/2 to pi/2 of 4cos^6(x) is (5pi)/4