数学题(有关不定积分运算)
-∫dcost)/(2-cos^2t)=-1/2√2*∫[1/(√2-cost)+1/(√2+cost)]d(cost)这一步怎么变来的?求详解~~谢谢大家了!!!!...
- ∫dcost)/(2 - cos^2 t)
= - 1/2√2 * ∫[ 1/(√2-cost)+1/(√2+cost) ]d(cost)
这一步怎么变来的?求详解~~谢谢大家了!!!! 展开
= - 1/2√2 * ∫[ 1/(√2-cost)+1/(√2+cost) ]d(cost)
这一步怎么变来的?求详解~~谢谢大家了!!!! 展开
2个回答
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1/(2 - cos^2 t)=1/(√2-cost)(√2+cost)=(1/(2√2) *2√2)/(√2-cost)(√2+cost)
=1/(2√2) *((√2-cost)+(√2+cost))/(√2-cost)(√2+cost)
=1/2√2 * [ 1/(√2-cost)+1/(√2+cost) ]
所以有:
- ∫dcost)/(2 - cos^2 t)
= - 1/2√2 * ∫[ 1/(√2-cost)+1/(√2+cost) ]d(cost)
=1/(2√2) *((√2-cost)+(√2+cost))/(√2-cost)(√2+cost)
=1/2√2 * [ 1/(√2-cost)+1/(√2+cost) ]
所以有:
- ∫dcost)/(2 - cos^2 t)
= - 1/2√2 * ∫[ 1/(√2-cost)+1/(√2+cost) ]d(cost)
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