作业帮高等数学三角函数不定积分图中的不定积分结果是多少啊?
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高等数学三角函数不定积分图中的不定积分结果是多少啊?
高等数学三角函数不定积分
图中的不定积分结果是多少啊?
高等数学三角函数不定积分图中的不定积分结果是多少啊?
如图
你令sinx的平方=t，运算两步，得到t的1/2和（1-t）的3/2次幂的多项式，不要再继续算t，将t在欢迎回sinx的平方。就得到一个叫容易算的式子了，在算几步就可以了，这步骤我写不上去，你试试，应该可以的
[sin(x)]^6+[cos(x)]^6=[sin(x)]^4+[cos(x)]^4-[sin(x)]^2*[cos(x)]^2=1-3*[sin(x)]^2*[cos(x)]^2=1-(3/4)*[sin(2x)]^2=1/4+(3/4)*[cos(2x)]^2
于是积分函数变为1/{1/4+(3/4)*[cos(2x)]^2}，分子分母同时除以[cos(2x)]^2得：积分函数=[s...
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[sin(x)]^6+[cos(x)]^6=[sin(x)]^4+[cos(x)]^4-[sin(x)]^2*[cos(x)]^2=1-3*[sin(x)]^2*[cos(x)]^2=1-(3/4)*[sin(2x)]^2=1/4+(3/4)*[cos(2x)]^2
于是积分函数变为1/{1/4+(3/4)*[cos(2x)]^2}，分子分母同时除以[cos(2x)]^2得：积分函数=[sec(2x)]^2/{(1/4)*[sec(2x)]^2+(3/4)}=[sec(2x)]^2/{[tan(2x)]^2+1}
原式=(1/2)*积分号:1/{[tan(2x)]^2+1}d[tan(2x)]=arctan{(1/2)*tan(2x)}+C
收起
[(cos(x))^4+(sin(x)^4)]
=[(cos(x))^4+(sin(x)^4)]*[(cos(x))^2+(sin(x)^2]
=(cos(x))^6 + (sin(x))^6 +(cos(x))^4 (sin(x))^2
+(sin(x))^4 (cos(x))^2
=(cos(x))^6 + (sin(x))^6 +(cos(x))^2 (sin...
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[(cos(x))^4+(sin(x)^4)]
=[(cos(x))^4+(sin(x)^4)]*[(cos(x))^2+(sin(x)^2]
=(cos(x))^6 + (sin(x))^6 +(cos(x))^4 (sin(x))^2
+(sin(x))^4 (cos(x))^2
=(cos(x))^6 + (sin(x))^6 +(cos(x))^2 (sin(x))^2 [(cos(x))^2+(sin(x))^2]
(cos(x))^6 + (sin(x))^6
= [(cos(x))^4+(sin(x)^4)]- (cos(x))^2 (sin(x))^2
=[(cos(x)^2 - (sin(x))^2]^2 +(sin(x))^2 (cos(x))^2
=(cos(2x))^2+(1/4)(sin(2x))^2
=[1+(1/4)(tan(2x)^2)]/(sec(2x)^2)
∫dx/[(sin(x))^6+(cos(x))^6]
= ∫(sec(2x))^2dx / [1+(1/4)(tan(2x))^2]
= ∫ (sec(2x))^2dx /
[1+{(1/2) tan(2x)}^2]
设z= (1/2) tan(2x)
∫dz/(1+z^2)=arctan(z) + C
∫ (sec(2x))^2dx /
[1+{(1/2) tan(2x)}^2]
=∫ d[(1/2) tan(2x)]/
[1+{(1/2) tan(2x)}^2]
=arctan[(1/2)tan(2x)] + C
收起
-tan^(-1)(2 cot(2 x))+c
过程太难输入了，保证正确