integral from pi to 0 of sin^4(θ)

解答

\int_{π}^{0} sin^{4} (θ) d θ

- \frac{3 π}{8}

十进制

- 1.17809 \dots

求解步骤

\int_{π}^{0} sin^{4} (θ) d θ

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{π} sin^{4} (θ) d θ

使用降次积分法

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

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

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

化简

- (- [\frac{cos ( θ ) sin ^{3} ( θ )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : [\frac{1}{4} sin^{3} (θ) cos (θ)]_{0}^{π} - \frac{3 π}{8}

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

带入上下限计算后得:

0

= 0 - \frac{3 π}{8}

化简

= - \frac{3 π}{8}

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