# Thread: Integral sin^2(x) + 2sin(t) +1

1. ## Integral sin^2(x) + 2sin(t) +1

How did people get the integral of 1 to be 3/2(x)?
I'm confused.

what is the integral of sin^2(x)?

2. Do you mean: $\displaystyle \int \left(\sin^{2} x + 2\sin x + 1 \right) dx$

First off, $\displaystyle \int 1 \: dx = \int 1x^{0} dx = \frac{x^{1}}{1} = x + c$

As for integrating $\displaystyle \sin^{2} x$, use the double angle identity: $\displaystyle \sin^{2} x = \frac{1 - \cos (2x)}{2}$

3. ## ok

I got 1/2x -1/4sin(2x)

how do I put this back in terms of sin(t) = x-1
(the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)

4. ## ok

I got 1/2x -1/4sin(2x)

how do I put this back in terms of sin(t) = x-1
(the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)

5. Originally Posted by khuezy
I got 1/2x -1/4sin(2x)

how do I put this back in terms of sin(t) = x-1
(the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)
You made the substitution $\displaystyle x = \sin\theta + 1,\text{ where }-\frac\pi2\leq\theta\leq\frac\pi2$, yes?

So, we have:

$\displaystyle \theta = \arcsin(x - 1)$,

$\displaystyle \sin2\theta = 2\sin\theta\cos\theta\text{\quad\color{red}(trig identity)}$,

and

$\displaystyle \cos\theta = \sqrt{\cos^2\theta} = \sqrt{1 - \sin^2\theta}\text{\color{red}\quad(since the cosine is positive for our values of theta)}$

$\displaystyle \Rightarrow\cos\theta = \sqrt{1 - (x - 1)^2}$

Now just make the appropriate substitutions.