∫ sin^6x cos^3x dx
=∫sin^6x (1-sin^2x) cosx dx
let,sinx=z
=d/dx sinx=dz/dx
=cosx dx=dz
so,∫ sin^6x cos^3x dx=∫sin^6x cosx dx-∫sin^6x sin^2x cosx dx
=∫z^6 dz-∫z^8 dz
=z^6+1/6+1-z^8+1/8+1 +c
=z^7/7+z^9/9+c
=sin^7x/7+sin^9x/9 +c is this right?

Question

∫ sin^6x cos^3x dx
=∫sin^6x (1-sin^2x) cosx dx
let,sinx=z
=>d/dx sinx=dz/dx
=>cosx dx=dz
so,∫ sin^6x cos^3x dx=∫sin^6x cosx dx-∫sin^6x sin^2x cosx dx
=∫z^6 dz-∫z^8 dz
=z^6+1/6+1-z^8+1/8+1 +c
=z^7/7+z^9/9+c
=sin^7x/7+sin^9x/9 +c is this right?

hartnn · Answer

yes! thats absolutely correct :)
good work!

hartnn · Answer

just that the sign in between the term should be minus

anonymous · Answer

Thank you very much :) genius sir :)

hartnn · Answer

=∫z^6 dz-∫z^8 dz
=z^6+1/6+1-z^8+1/8+1 +c
=z^7/7-z^9/9+c
=sin^7x/7-sin^9x/9 +c

anonymous · Answer

oh sorry my head is hot little :)

hartnn · Answer

no problem :) welcome ^_^

anonymous · Answer

^_^ thank you again