Draw the graph of `y=sinxcos^(2)x`

Explanation

We have `y=f(x)=sinxcos^(2)x`
Clearly, the domain of `f(x)` is R.
Also the period of `f(x)` is R.
Let us first draw the graph of `y=f(x)` for `0 lex le2pi`
`f(x) =0 therefore sinx=0` or `cos^(2)x=0`
`therefore x=0, pi, 2pi` or `x=pi//2, 3pi//2`
`f(x) gt 0`, for `x in (0,pi)` and `f(x) lt 0`, for `x in (pi, 2pi)`
Now `f^(')(x) = cosx cos^(2)x + sinx(-2sinxcosx)`
`=cosx(cos^(2)x-2sin^(2)x)`
`f^(')(x)=0 therefore cosx=0` or `tan^(2)x=1//2`
`therefore x=pi/2, (3pi)/2` or `x=tan^(-1)1/sqrt(2), pi-tan^(-1) 1/sqrt(2), pi + tan^(-1)1/sqrt(2) , 2pi -tan^(-1)1/sqrt(2)`
Now we can draw the graph of `y=f(x)` for `x in (0,pi)`
For `x in (pi,2pi)`, we reflect the graph of `y=f(x)` for `x in (0,pi)` in the x-axis as `f(pi-x) = -f(pi+x)`
`f(tan^(-1)1/sqrt(2))=f(pi-tan^(-1)1/sqrt(2))=2/(3sqrt(3))`
From the above discussion, we have, following important points on graph paper.

`f(x)` increases from O to A, decreases from A to B, increases from B to C, decreases from C to D.
Hence the graph of `y=f(x)` for ` 0 lt x lt pi` is as shown in the following figure.

For `pi lt x lt 2pi`, the graph of `y=f(x)` lies below the x-axis.
Hence, the graph of `y=f(x)` for `pi lt x lt 2pi` is as shown in the following figure.

Since the period of `f(x)` is `2pi`, the graph for R is as follows: