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

The integral from 0 to pi/2 of sin^4(θ) is (3pi)/(16)

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integral from 0 to pi/2 of sin^4(θ)

\int_{0}^{\frac{π}{2}} sin^{4} (θ) d θ

\frac{3 π}{16}

Decimal

0.58904 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin^{4} (θ) d θ

Apply Integral Reduction

= - [\frac{cos ( θ ) sin ^{3} ( θ )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (θ) d θ

\int_{0}^{\frac{π}{2}} sin^{2} (θ) d θ = \frac{π}{4}

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

Simplify

- [\frac{cos ( θ ) sin ^{3} ( θ )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4} : - [\frac{1}{4} sin^{3} (θ) cos (θ)]_{0}^{\frac{π}{2}} + \frac{3 π}{16}

= - [\frac{1}{4} sin^{3} (θ) cos (θ)]_{0}^{\frac{π}{2}} + \frac{3 π}{16}

Compute the boundaries:

0

= - 0 + \frac{3 π}{16}

Simplify

= \frac{3 π}{16}

\int_{1}^{2} \frac{x}{( x ^{2} + 2 ) ^{2}} d x

\int_{1}^{2} (x^{3} + 3 x^{2} - 4 x) d x

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

\int_{0}^{0.5} 1 d x

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

What is the integral from 0 to pi/2 of sin^4(θ) ?
The integral from 0 to pi/2 of sin^4(θ) is (3pi)/(16)