高数高手请进,求微积分∫1/1+sinx dx∫(arctan1/x)/1+x^2 dx∫x^3/(1+x^2)^(1/2) dx∫ln(tanx)/sin(cosx) dx分太少了,但是已经倾其所有啦,很不好意思请大家帮帮忙
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/26 02:49:46
![高数高手请进,求微积分∫1/1+sinx dx∫(arctan1/x)/1+x^2 dx∫x^3/(1+x^2)^(1/2) dx∫ln(tanx)/sin(cosx) dx分太少了,但是已经倾其所有啦,很不好意思请大家帮帮忙](/uploads/image/z/6802125-69-5.jpg?t=%E9%AB%98%E6%95%B0%E9%AB%98%E6%89%8B%E8%AF%B7%E8%BF%9B%2C%E6%B1%82%E5%BE%AE%E7%A7%AF%E5%88%86%E2%88%AB1%2F1%2Bsinx+dx%E2%88%AB%28arctan1%2Fx%29%2F1%2Bx%5E2+dx%E2%88%ABx%5E3%2F%281%2Bx%5E2%29%5E%281%2F2%29+dx%E2%88%ABln%28tanx%29%2Fsin%28cosx%29+dx%E5%88%86%E5%A4%AA%E5%B0%91%E4%BA%86%2C%E4%BD%86%E6%98%AF%E5%B7%B2%E7%BB%8F%E5%80%BE%E5%85%B6%E6%89%80%E6%9C%89%E5%95%A6%2C%E5%BE%88%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%E8%AF%B7%E5%A4%A7%E5%AE%B6%E5%B8%AE%E5%B8%AE%E5%BF%99)
高数高手请进,求微积分∫1/1+sinx dx∫(arctan1/x)/1+x^2 dx∫x^3/(1+x^2)^(1/2) dx∫ln(tanx)/sin(cosx) dx分太少了,但是已经倾其所有啦,很不好意思请大家帮帮忙
高数高手请进,求微积分
∫1/1+sinx dx
∫(arctan1/x)/1+x^2 dx
∫x^3/(1+x^2)^(1/2) dx
∫ln(tanx)/sin(cosx) dx
分太少了,但是已经倾其所有啦,很不好意思请大家帮帮忙
高数高手请进,求微积分∫1/1+sinx dx∫(arctan1/x)/1+x^2 dx∫x^3/(1+x^2)^(1/2) dx∫ln(tanx)/sin(cosx) dx分太少了,但是已经倾其所有啦,很不好意思请大家帮帮忙
第一个:∫1/1+sinx dx
万能代换 t=tan(x/2),
1/(1+sinx)dx=1/(1+2t/(1+t^2))*2t/(1+t^2)dt=2t/(1+t)^2dt=(2/(1+t)-2/(1+t)^2)dt=2ln(1+t)+2/(1+t)+C=2ln(1+tan(x/2)+2/(1+tan(x/2))+C
第二个:∫(arctan1/x)/1+x^2 dx
∫[arctan(1/x)]dx/(1+x²)
=-∫[x²arctan(1/x)]d(1/x)/(1+x²)
=-∫[arctan(1/x)]d(1/x)/(1+1/x²)
=-∫[arctan(1/x)]darctan(1/x)
=-1/2[arctan(1/x)]^2+c
第三个:
∫x^2 * arctanxdx
=1/3×∫arctanxd(x^3)
=1/3×x^3×arctanx-1/3×∫x^3/(1+x^2)
=1/3×x^3×arctanx-1/3×∫(x^3+x-x)/(1+x^2)
=1/3×x^3×arctanx-1/3×∫xdx+1/3×∫x/(1+x^2)
=1/3×x^3×arctanx-1/6×x^2+1/6×ln(1+x^2)+C
第四个:
万能代换 t=tan(x/2),
1/(1+sinx)dx=1/(1+2t/(1+t^2))*2t/(1+t^2)dt=2t/(1+t)^2dt=(2/(1+t)-2/(1+t)^2)dt=2ln(1+t)+2/(1+t)+C=2ln(1+tan(x/2)+2/(1+tan(x/2))+C
第二个:∫(arctan1/x)/1+x^2 dx
∫[arctan(1/x)]...
全部展开
万能代换 t=tan(x/2),
1/(1+sinx)dx=1/(1+2t/(1+t^2))*2t/(1+t^2)dt=2t/(1+t)^2dt=(2/(1+t)-2/(1+t)^2)dt=2ln(1+t)+2/(1+t)+C=2ln(1+tan(x/2)+2/(1+tan(x/2))+C
第二个:∫(arctan1/x)/1+x^2 dx
∫[arctan(1/x)]dx/(1+x²)
=-∫[x²arctan(1/x)]d(1/x)/(1+x²)
=-∫[arctan(1/x)]d(1/x)/(1+1/x²)
=-∫[arctan(1/x)]darctan(1/x)
=-1/2[arctan(1/x)]^2+c
第三个:
∫x^2 * arctanxdx
=1/3×∫arctanxd(x^3)
=1/3×x^3×arctanx-1/3×∫x^3/(1+x^2)
=1/3×x^3×arctanx-1/3×∫(x^3+x-x)/(1+x^2)
=1/3×x^3×arctanx-1/3×∫xdx+1/3×∫x/(1+x^2)
=1/3×x^3×arctanx-1/6×x^2+1/6×ln(1+x^2)+C
收起
我用另一种的方法来做第一题吧 原式=∫(1-sinx)dx/(cosx)^2=∫(secx)^2+∫d(cosx)/(cosx)^2=tanx-secx+c这个好像还行,