Indefinite Integral (1 Viewer)

[juantheron]

If , Then

 

[deswa1]

<img src="http://latex.codecogs.com/gif.latex?7I_{4,3}-4I_{3,2}=7\int cos4xsin3x dx-4\int cos3xsin2xdx\\ =\frac{7}{2}\int 2cos4xsin3xdx-2\int 2cos3xsin2xdx\\ =\frac{7}{2}\int sin7x-sinxdx-2\int sin5x-sinxdx\\ =\frac{7}{2}(\frac{-cos7x}{7}+cosx)-2(\frac{-cos5x}{5}+cosx) +C\\ =\frac{-cos7x}{2}+\frac{2cos5x}{5}+\frac{3cosx}{2}+C" title="7I_{4,3}-4I_{3,2}=7\int cos4xsin3x dx-4\int cos3xsin2xdx\\ =\frac{7}{2}\int 2cos4xsin3xdx-2\int 2cos3xsin2xdx\\ =\frac{7}{2}\int sin7x-sinxdx-2\int sin5x-sinxdx\\ =\frac{7}{2}(\frac{-cos7x}{7}+cosx)-2(\frac{-cos5x}{5}+cosx) +C\\ =\frac{-cos7x}{2}+\frac{2cos5x}{5}+\frac{3cosx}{2}+C" />

 

[barbernator]

<img src="http://latex.codecogs.com/gif.latex?7I_{4,3}-4I_{3,2}=7\int cos4xsin3x dx-4\int cos3xsin2xdx\\ =\frac{7}{2}\int 2cos4xsin3xdx-2\int 2cos3xsin2xdx\\ =\frac{7}{2}\int sin7x-sinxdx-2\int sin5x-sinxdx\\ =\frac{7}{2}(\frac{-cos7x}{7}+cosx)-2(\frac{-cos5x}{5}+cosx) +C\\ =\frac{-cos7x}{2}+\frac{2cos5x}{5}+\frac{3cosx}{2}+C" title="7I_{4,3}-4I_{3,2}=7\int cos4xsin3x dx-4\int cos3xsin2xdx\\ =\frac{7}{2}\int 2cos4xsin3xdx-2\int 2cos3xsin2xdx\\ =\frac{7}{2}\int sin7x-sinxdx-2\int sin5x-sinxdx\\ =\frac{7}{2}(\frac{-cos7x}{7}+cosx)-2(\frac{-cos5x}{5}+cosx) +C\\ =\frac{-cos7x}{2}+\frac{2cos5x}{5}+\frac{3cosx}{2}+C" />

moral of the story, learn product sums and differences of trig

 

[deswa1]

moral of the story, learn product sums and differences of trig

This. Even for 3U students, it can come in handy sometimes and its not really that hard...