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

The integral from-pi/2 to pi/2 of 6cos^4(x) is (9pi)/4

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

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

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

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

= 6 \cdot \int_{- \frac{π}{2}}^{\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]_{- \frac{π}{2}}^{\frac{π}{2}}

= 6 [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{- \frac{π}{2}}^{\frac{π}{2}}

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

= 6 [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{- \frac{π}{2}}^{\frac{π}{2}}

Simplify

6 [cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{- \frac{π}{2}}^{\frac{π}{2}} : 6 [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{- \frac{π}{2}}^{\frac{π}{2}}

= 6 [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{- \frac{π}{2}}^{\frac{π}{2}}

Compute the boundaries:

\frac{3 π}{8}

= 6 \cdot \frac{3 π}{8}

Simplify

= \frac{9 π}{4}

\int_{1}^{7} (\frac{4}{x}) d x

\int_{2}^{6} \frac{1}{5} x^{2} d x

\int_{0}^{5} x^{2} - 3 x - 4 d x

\int_{1}^{3} 3 x^{2} - 2 x + 1 d x

\int_{- 1}^{1} \frac{x ^{3}}{x ^{2} + 1} d x

What is the integral from-pi/2 to pi/2 of 6cos^4(x) ?
The integral from-pi/2 to pi/2 of 6cos^4(x) is (9pi)/4