Evaluate : (i) `intcos2x\ dx` (ii) `inte^((5x+3))dx` (iii) `int sec^(2)(3x+5)dx` (iv) `intsin^(3)xdx`

Explanation

(i) Put `2x="t so that "2dx=dt" or "=(1)/(2)dt`.
`:." "intcos2x=(1)/(2)intcosdt=(1)/(2)sint+C=(1)/(2)sin2x+C`.
(ii) Put (5x+3)=t so that 5dx=dt or dx `=(1)/(5)dt`
`:." "inte^((5x+3))dx=(1)/(5)*e^(t)+C=(1)/(5)e^((5x+3))+C`.
(iii) Put (3x+5)dx=t so that 3 dx = dt or `dx=(1)/(3)dt`.
`:.intsec^(2)(3x+5)dx=(1)/(3)=intsec^(2)t" dt"=(1)/(3)tant+C=(1)/(3)tan(3x+5)+C`.
(iv) We know that `sin3x=3 sin x-4sin^(3)x`.
`:." "sin^(3)x=(1)/(4)(3sinx-sin3x)`.
so, `intsin^(3)xdx=int((3)/(4)sinx-(1)/(4)sin3x)dx=(3)/(4)intsinxdx-(1)/(4)fsin3xdx=(3)/(4)intxdx-(1)/(4)(-cosx)-(1)/(4)*(-cos3x)/(3)+C=-(3)/(4)cosx+(cos3x)/(12)+C`.