What is the integral from 0 to pi of cos^4(θ) ?

The integral from 0 to pi of cos^4(θ) is (3pi)/8

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integral from 0 to pi of cos^4(θ)

\int_{0}^{π} cos^{4} (θ) d θ

\frac{3 π}{8}

Decimal

1.17809 \dots

Solution steps

\int_{0}^{π} cos^{4} (θ) d θ

Simplify

cos^{4} (θ) : cos^{3} (θ) cos (θ)

= \int_{0}^{π} 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]_{0}^{π}

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{3 π}{8}

= \frac{3 π}{8}

\int_{- 1}^{1} (x^{3} - x^{4}) d x

\int_{0}^{1} 3 x^{2} arcsin (x) d x

\int_{- 5}^{2} \frac{5}{x ^{4}} d x

\int_{0}^{2} - \frac{3}{2} x + 3 d x

\int_{1}^{3} (1 + x) d x

What is the integral from 0 to pi of cos^4(θ) ?
The integral from 0 to pi of cos^4(θ) is (3pi)/8