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

The integral from-pi/2 to pi/2 of 4cos^4(θ) is (3pi)/2

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

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

\frac{3 π}{2}

Decimal

4.71238 \dots

Solution steps

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

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^{4} (θ) d θ

Simplify

cos^{4} (θ) : cos^{3} (θ) cos (θ)

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

Apply Integration By Parts

: [cos^{3} (θ) sin (θ) - \int - 3 cos^{2} (θ) sin^{2} (θ) d θ]_{- \frac{π}{2}}^{\frac{π}{2}}

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{3 π}{8}

= 4 \cdot \frac{3 π}{8}

Simplify

= \frac{3 π}{2}

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