主要内容:
本题通过三角函数恒等变形和三角函数换元法两种方法,介绍计算定积分∫dx/[√2+sin(x+1)+cos(x+1)]的方法和步骤,并可以观察出,同一个不定积分结果的表达式可以不唯一。
.三角函数恒等变形法
I=∫dx/[√2+sin(x+1)+cos(x+1)],
根据公式sin(x+π/4)=sinxcosπ/4+cosxsinπ/4变形为:
I==∫dx/{√2+√2[sin(x+1)cosπ/4+cos(x+1)sinπ/4]}
=∫dx/[√2+√2sin(x+1+π/4)],以下提取公因数系数,
=(1/√2)∫dx/[1+sin(x+1+π/4)],以下根据sin^2x+cosx^2=1变形为,
=(√2/2)∫dx/[sin(1/2)(x+1+π/4)+cos(1/2)(x+1+π/4)]^2,
=(√2/2)∫dx/{√2sin[(1/2)(x+1+π/4)+π/4]}^2
=(√2/4)∫dx/sin^2[(1/2)(x+1)+3π/8],以下根据公式cscx=1/sinx变形为,
=(√2/4)∫csc^2[(1/2)(x+1)+3π/8]dx,以下对微分微元dx进行变形,
=(√2/2)∫csc^2[(1/2)(x+1)+3π/8]d[(1/2)(x+1)],
以下有积分公式∫csc^2xdx=-cotx+C变形得,
I=-(√2/2)cot[(1/2)(x+1)+3π/8]+C。
.三角函数换元法
设tan(1/2)(x+1)=t,则x=(2arctant-1),
同时由三角万能公式有:
sin(x+1)=2t/(1+t^2),cos(x+1)=(1-t^2)/(1+t^2),
代入所求不定积分,则:
I=∫dx/√2+sin(x+1)+cos(x+1),
=∫d[(2arctant-1)]/√2+2t/(1+t^2)+(1-t^2)/(1+t^2),
=2∫[1/(1+t^2)]dt/{[√2(1+t^2)+2t+(1-t^2)]/(1+t^2)},
=2∫dt/[√2(1+t^2)+2t+(1-t^2)],
以下对分母进行关于t的二次函数变形为,
I=2∫dt/[(√2-1)(t+√2+1)^2]
=2(√2+1)∫dt/(t+√2+1)^2,
以下根据不定积分公式∫dx/x^2=-1/x+C计算得,
代入t=tan(1/2)(x+1),即可计算出本题不定积分结果为,
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1/(1 - cos2x)的不定积分 - : ∫ dx/cos2x= ∫ sec2x dx= ∫ sec2x * (sec2x+tan2x)/(sec2x+tan2x) dx= (1/2)∫ (sec²2x+sec2xtan2x)/(sec2x+tan2x) d(2x)= (1/2)∫ d(tan2x+sec2x)/(sec2x+tan2x)= (1/2)ln|sec2x+tan2x| + c
1/cos2x的不定积分 - : |∫= ∫ sec2x dx= ∫ sec2x * (sec2x+tan2x)/(sec2x+tan2x) dx= (1/2)∫ (sec²2x+sec2xtan2x)/(sec2x+tan2x) d(2x)= (1/2)∫ d(tan2x+sec2x)/(sec2x+tan2x)= (1/2)ln|sec2x+tan2x| + C
1/(1+cos2x)不定积分是多少 - : 1/2*tanx +c
1 - cos2x的原函数 - : cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x 所以1-cos2x=1-(1-2sin²x)=2sin²x如果你这个是导数,那么sin2x求导是2cos2x,所以1-cos2x 的原函数是 -1/2*sin2x + x + C
1 - cos2x在区间0 - 排的定积分是多少 - : 1-cos2x=2(sinx)^2在0到π上sinx恒为正从而简化为根号2*sinx积分为根号2*(-cos(π)+cos0)=2根号2
∫√1 - cos2x dx,积分从0到2π - : ∫√(1-cos2x)dx=∫√2|sinx|dx=2√2∫sinxdx=-2√2cosx|=4√2.