三角函数的积分如何计算?

答：\[ \int \csc(x) \, dx = -\ln|\csc(x) + \cot(x)| + C \]5. **反正弦、反余弦等**：\[ \int \arcsin(x) \, dx = x\sqrt{1-x^2} + C \]\[ \int \arccos(x) \, dx = x\sqrt{1-x^2} - C \]\[ \int \arctan(x) \, dx = \frac{x}{2} \ln\le...

2024-05-19 回答者: zhuangmtuo3 3个回答

不定积分∫√(a^2+ x^2) dx求值?

答：将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数）

2024-05-20 回答者: 风林网络手游平台 1个回答

三角函数的积分如何求

答：1. 反正弦函数：$\int \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1 - x^2} + C 2. 反余弦函数：$\int \arccos(x) \, dx = x \arccos(x) - \sqrt{1 - x^2} + C 3. 反正切函数：$\int \arctan(x) \, dx = x \arctan(x) - \frac{1}{2} \ln(1 + x^2...

2024-05-19 回答者: 淘无忧 1个回答

∫√( a^2- x^2) dx=什么?

答：将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数）

2024-05-20 回答者: 风林网络手游平台 1个回答

不定积分∫√( a^2- x^2) dx=?

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答

如何由不定积分求值?

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答

已知函数f(x)= asint(a^2- x^2),求

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答

∫(a^2- x^2) dx怎样积分

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答

如何计算不定积分∫√( a^2- x^2) dx

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答

求不定积分∫√( a^2- x^2)的值域?

答：a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数...

2024-05-20 回答者: 风林网络手游平台 1个回答