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受欢迎的微积分 >

integral of sin^5(2θ)

解答

\int sin^{5} (2 θ) d θ

解答

32 (- \frac{cos ^{6} ( θ )}{6} + \frac{cos ^{8} ( θ )}{4} - \frac{cos ^{10} ( θ )}{10}) + C

求解步骤

\int sin^{5} (2 θ) d θ

使用三角恒等式改写

= \int 32 cos^{5} (θ) sin^{5} (θ) d θ

提出常数:

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

= 32 \cdot \int cos^{5} (θ) sin^{5} (θ) d θ

使用三角恒等式改写

= 32 \cdot \int (1 - cos^{2} (θ))^{2} sin (θ) cos^{5} (θ) d θ

使用换元积分法

= 32 \cdot \int - u^{5} (1 - u^{2})^{2} d u

乘开

- u^{5} (1 - u^{2})^{2} : - u^{5} + 2 u^{7} - u^{9}

= 32 \cdot \int - u^{5} + 2 u^{7} - u^{9} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 32 (- \int u^{5} d u + \int 2 u^{7} d u - \int u^{9} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int 2 u^{7} d u = \frac{u ^{8}}{4}

\int u^{9} d u = \frac{u ^{10}}{10}

= 32 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{4} - \frac{u ^{10}}{10})

u = cos (θ)

代回

= 32 (- \frac{cos ^{6} ( θ )}{6} + \frac{cos ^{8} ( θ )}{4} - \frac{cos ^{10} ( θ )}{10})

解答补常数

= 32 (- \frac{cos ^{6} ( θ )}{6} + \frac{cos ^{8} ( θ )}{4} - \frac{cos ^{10} ( θ )}{10}) + C

作图

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