∫sin3xdx=∫sin2x⋅sinxdx=replacecosx=td(cosx)=dt−sinxdx=dtsin2x=1−cos2x=−∫(1−t2)dt=−(t−t33+C)==−(cosx−cos3x3+C)\int sin^3x dx=\int sin^2 x\cdot sinx dx=\\ replace\\ cos x=t\\ d(cosx)=dt\\ -sinxdx=dt\\ sin^2x=1-cos^2x\\ =-\int(1-t^2)dt=-(t-\frac{t^3}{3}+C)=\\ =-(cosx-\frac{cos^3x}{3}+C)∫sin3xdx=∫sin2x⋅sinxdx=replacecosx=td(cosx)=dt−sinxdx=dtsin2x=1−cos2x=−∫(1−t2)dt=−(t−3t3+C)==−(cosx−3cos3x+C)
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