Question

Integrate:\(\int\limits_{-1}^{1}(1+\sqrt{1-x^2})dx\)

Answer 1

I tried using trig substitution but got lost halfway..

Answer 2

Actually if you graph this function, you'll know the answer without even finding an anti-derivative.

Answer 3

But, the question doesn't want me to graph it :\

Answer 4

But anyway, as for \[
\sqrt{1-x^2}~dx
\]You would set \(x=\sin t\) \[
\sqrt{1-\sin^2 t} ~d(\sin t) = \sqrt{\cos^2 t} (\cos t)~dt=|\cos t| \cos t~dt
\]

Answer 5

\(x+\int\limits_{-1}^{1}(\sqrt{1-sin^2t})*cost dx\)
\(x+\int\limits_{x=-1}^{x=1}(cost)*cost dt\)
\(x+\int\limits_{x=-1}^{x=1}(cos^2t dt\)
cos^2t=(1+cos2t)/2
\(x+\int\limits_{x=-1}^{x=1}(\frac{1+cos^2t}{2} *dt\)

Answer 6

This is what i did

Answer 7

The limits \(x\in (-1,1)\) translate to \(t\in (-\pi/2,\pi,2\), and so \(|cos t| = \cos t\).

Answer 8

Right, right

Answer 9

Why do you have the square term?

Answer 10

correction: \(x+\int\limits_{-\pi/2}^{\pi/2}(\frac{1+cos(2t)}{2}\)
sorry about that

Answer 11

Okay, I'm still confused though. Why do you have an \(x\)? The variable of integration goes away when you integrate.

Answer 12

i was thinking of antiderivative of 1 be x?

Answer 13

\[
\int_{-1}^1dx = 2
\]

Answer 14

Anyway, I think you can do the rest

Answer 15

I should have applied FTOC before, and consider it more xD

Answer 16

?

Answer 17

\[2+\frac{1}{2}\int\limits_{-\pi/2}^{\pi/2}1+\cos(2t) dt\]