Question

Question #eb58b

Answer 1

`intxsinx^2dx=-1/2cosu+c`

Explanation:

`intxsinx^2dx`
`int(sinx^2)xdx`

By the method of substitution
Let
`x^2=u`
Differentiating wrt x
`2x=du/dx`
Separating the variables
`2xdx=du`
Simplifying
`xdx=1/2(du)/dx`
Thus,
`int(sinx^2)xdx=intsinu(1/2du)`
`int1/2sinudu`
`1/2intsinudu`
`1/2(-cosu)+c`
`-1/2cosu+c`

Now,

`intxsinx^2dx=-1/2cosu+c`

Answer 2

`-cos(x^2)/2+C`

Explanation:

Okay, we have `intxsin(x^2)dx`.

We can use integration by substitution. To do so, we must rewrite the equation as such:

`intf(g(x))g'(x)dx`.

In our case, `f(u)=sinu` and `g(x)=x^2`. However, `g'(x)=2x`.

So we can write this as `1/2intsin(x^2)xdx`.

As `xdx` becomes `du`, and `sin(x^2)` becomes `u`, we get:

`1/2intsinudu`

`1/2*-cosu`

`-cosu/2`.

But remember, `u=x^2`. So input:

`-cos(x^2)/2`

Then add the constant of integration:

`-cos(x^2)/2+C`

Answer 3

`intxsinx^2dx=-1/2cosx^2+c`

Explanation:

Another approach is by inspection

it is known that by the chain rule

`d/(dx)(cosf(x))=-f'(x)sinf(x)`

so consider
`d/(dx)(cosx^2)=-2xsinx^2`

`:. intxsinx^2dx=-1/2cosx^2+c`