Each part is a standard trig form (sin, cos, cosec2) evaluated at a linear argument ax+b; integrate as usual and divide by a.
Step 1. Part (i). a=3; ∫sinxdx=−cosx+c, so
∫sin3xdx=−31cos3x+c.
Step 2. Part (ii). 5−11x has a=−11; ∫cosxdx=sinx+c, so
∫cos(5−11x)dx=−111sin(5−11x)+c=−111sin(5−11x)+c.
Step 3. Part (iii). a=5; ∫cosec2xdx=−cotx+c, so
∫cosec2(5x−7)dx=51(−cot(5x−7))+c=−51cot(5x−7)+c.
Step 4. Check by differentiating. dxd(−31cos3x)=31sin3x⋅3=sin3x ✓. dxd(−111sin(5−11x))=−111cos(5−11x)⋅(−11)=cos(5−11x) ✓. dxd(−51cot(5x−7))=51cosec2(5x−7)⋅5=cosec2(5x−7) ✓.
✓Final answer
(i) −31cos3x+c (ii) −111sin(5−11x)+c (iii) −51cot(5x−7)+c