7.1 Integration By Parts/29: Difference between revisions
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(Created page with "<math> f'(x)= \int_{}^{}\cos(x)ln(sin(x))\cdot dx </math> <br><br> <math>\int_{}^{}\cos(x)ln(sin(x)\cdot dx=\int_{}^{}\ln(u)\cdot du</math><br> <math>u=sin(x)</math> <br> <math> du=cos(x)dx</math> <br><br> <math>=u \cdot ln(u)-u+c</math><br><br> <math>\text{Therefore, } f(x)=sin(x)ln(sin(x))-sin(x)+c</math>") |
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<math> f'(x)= \int_{}^{}\cos(x)ln(sin(x))\cdot dx </math> <br><br> | <math> f'(x)= \int_{}^{}\cos(x)\ln(sin(x))\cdot dx </math> <br><br> | ||
<math>\int_{}^{}\cos(x)ln(sin(x)\cdot dx=\int_{}^{}\ln(u)\cdot du</math><br> | <math>\int_{}^{}\cos(x)\ln(\sin(x))\cdot dx=\int_{}^{}\ln(u)\cdot du</math><br> | ||
<math>u=sin(x)</math> <br> <math> du=cos(x)dx</math> <br><br> | <math>u=\sin(x)</math> <br> <math> du=\cos(x)dx</math> <br><br> | ||
<math>=u \cdot ln(u)-u+c</math><br><br> | <math>=u \cdot ln(u)-u+c</math><br><br> | ||
<math>\text{Therefore, } f(x)=sin(x)ln(sin(x))-sin(x)+c</math> | <math>\text{Therefore, } f(x)=sin(x)ln(sin(x))-sin(x)+c</math> | ||
Revision as of 20:56, 26 November 2022
