how to solve the following integral?

Question

(1+sin2x) dx

the sin 2x should be sin squared x

Answer 1

you need to use the half angle formula:

sin(x)^2 = 1/2(1-cos(2x))

so you get:

1+1/2-1/2cos(2x) dx

3/2-1/2cos(2x) dx

let u = 2x, then du = 2dx and dx = du/2

3/2-1/2cos(u)du/2

3/2-cos(u)du/4

3/2x-sin(u)/4+c

3/2x-sin(2x)/4+c

Answer 2

if you see the square of a sine or a cosine, then you can probably reduce it by using a double angle formula:

remember that cos(2x) = cos^2(x) - sin^2(x) = 1 - 2 sin^2(x)
which means sin^2(x) = 1/2 - 1/2 cos(2x)

So now we can write the integral as:
integral(1 + 1/2 - 1/2cos(2x))

and if we use the chain rule, we can correctly guess that the answer is:
3/2 x - 1/4 sin(2x) + C

Answer 3

Integral ( ( 1 + sin^2(x) ) dx )

First off, you can separate the 1 and integrate that by itself. This gives us

Integral(1 dx) + Integral ( sin^2(x) dx )

x + Integral ( sin^2(x) dx )

Now, use the half angle formula identity.
sin^2(x) = (1/2)(1 - cos(2x)).

x + Integral ( (1/2)(1 - cos(2x)) dx )

Factor out (1/2), to obtain

x + (1/2) Integral ( (1 - cos(2x)) dx )

And now, this is an easy integral, as the argument inside the cosine is a linear one.

x + (1/2) ( x - (1/2)sin(2x) ) + C

x + (1/2)x - (1/4)sin(2x) + C

(3/2)x - (1/4)sin(2x) + C

Answer 4

integral 1dx+integral sin ^2 x dx
=?