作业帮1/(cosx+sinx)dx 积分怎样计算?还有e^-x^2dx积分谢谢
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/06 03:46:21
1/(cosx+sinx)dx 积分怎样计算?还有e^-x^2dx积分谢谢
1/(cosx+sinx)dx
e^-x^2dx
1/(cosx+sinx)dx
e^-x^2dx
∫1/(cosx+sinx)dx
=∫(cosx-sinx)dx/(cos2x)
=∫cosxdx/cos2x-∫sinxdx/cos2x
=∫dsinx/[1-2(sinx)^2]+∫dcosx/[2(cosx)^2-1]
设sinx=m,cosx=n
那么原式等于∫dm/(1+√2m)(1-√2m)+∫dn/(√2n+1)(√2n-1)
=1/2∫dm/(1+√2m)+1/2∫dm/(1-√2m)+1/2∫dn/(1+√2n)-1/2∫dn/(1-√2n)
=∫dx/(1+√2x)
=√2∫dx/(x+√2/2)
=√2d(x+√2/2)/(x+√2/2)
=√2ln(x+√2/2).
=∫(cosx-sinx)dx/(cos2x)
=∫cosxdx/cos2x-∫sinxdx/cos2x
=∫dsinx/[1-2(sinx)^2]+∫dcosx/[2(cosx)^2-1]
设sinx=m,cosx=n
那么原式等于∫dm/(1+√2m)(1-√2m)+∫dn/(√2n+1)(√2n-1)
=1/2∫dm/(1+√2m)+1/2∫dm/(1-√2m)+1/2∫dn/(1+√2n)-1/2∫dn/(1-√2n)
=∫dx/(1+√2x)
=√2∫dx/(x+√2/2)
=√2d(x+√2/2)/(x+√2/2)
=√2ln(x+√2/2).
1/(cosx+sinx)dx 积分怎样计算?还有e^-x^2dx积分谢谢
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