Question

How do you integrate `int (sinx +cosx)^2 dx`?

Answer 1

`=x+sin^2(x)+C`

Explanation:

`int\ (sin(x)+cos(x))^2\ dx`
`=int\ sin^2(x)+cos^2(x)+2sin(x)cos(x)\ dx` {Expanding}
`=int\ 1+2sin(x)cos(x)\ dx` {Since `sin^2(x)+cos^2(x)=1`}
`=int\1\ dx+2int\ sin(x)cos(x)\ dx` {Linearity of integration}
`=x+C+2int\ sin(x)cos(x)\ dx`

Now, substitue `u=sin(x)`, or `dx=(du)/cos(x)`

`=x+C+2int\ (ucos(x))/cos(x)\ du`
`=x+C+2int\ u\ du`
`=x+C+(2u^2)/2`
`=x+C+u^2`
`=x+sin^2(x)+C`

Answer 2

The antiderivative is `sin^2x + x + C`.

Explanation:

Here's another way of doing this. We expand, calling the integral `I`:

`I = int (sinx + cosx)^2 dx`

`I = int sin^2x + cos^2x+ 2sinxcosx dx`

`I = int 1 + 2sinxcosx`

`I = int 1 dx + int 2sinxcosxdx`

We now use the identity `sin2x= 2sinxcosx`.

`I = int 1 dx + int sin2xdx`

Now let `u = 2x`. Then `du = 2dx` and `dx= 1/2du`.

`I = int 1 dx + 1/2int sinu du`

`I = x - 1/2cosu`

`I = x - 1/2cos(2x) + C`

Now use `cos(2x) = 1 - 2sin^2x`.

`I = x - 1/2(1 - 2sin^2x) + C`

`I = x + 1/2 + sin^2x + C`

However, since `C` is included, we can just forget about the `1/2`.

`I = sin^2x+ x + C`

Hopefully this helps!