Find the value of the integral $\int_0 ^\sqrt2 \sqrt{4-x^2} \ dx$ Find the value of the integral $\int_0 ^\sqrt2 \sqrt{4-x^2} \ dx$
I was unsure how to do this question so I looked at the mark scheme, and it said use $x=2\sin\theta$ and so $dx=2\cos\theta \ d\theta$
Firstly, how do you know you need to use this substitution? What do you need to look out for?
I tried doing a bit more and got $\int_0 ^{\pi/4} \sqrt{4-4\sin^2 \theta}$ which I changed into $\sqrt{4-4+\cos^2 \theta}$ using the Pythagorean identity. The markscheme however then has $\int 2\cos\theta \times 2\cos\theta$. I am not sure where they got the first 2 from, but I know the second one is from the substitution
edit: I see the problem now, really careless algebraic mistake... 
 A: your problem is where you changed $(4-4\sin^2\theta)$ to $(4-4+\cos^2\theta)$. You should be changing it to $4(1-\sin^2\theta)=4(\cos^2\theta)$. The reason they know to use that substitution is because it's exactly the coefficient (2) squared that will match with the 4 to factor out, leaving you with $1-\sin^2\theta=\cos^2\theta$. Then you'll have
$\int_0^\sqrt2 \sqrt{4\cos^2\theta}(2\cos\theta d\theta)$
$ = \int_0^\sqrt2 (2\cos\theta\times 2\cos\theta)\:d\theta$
$ =\int_0^\sqrt2 4\cos^2\theta \:d\theta$
A: if you interpret the integral $ \int_0^{\sqrt2}\sqrt{4-x^2}\, dx$ as the area of the region in the first quadrant bounded by $x = 0, y=0, x = \sqrt 2$ and $y^2+x^2 = 4,$  then the area is the sum of an isosceles right ange triangle with base $\sqrt 2$ and an eighth of a circular sector of radius. therefore $$ \int_0^{\sqrt2}\sqrt{4-x^2}\, dx = \frac 12 \sqrt 2 \sqrt 2 + \frac 18 \pi( 2)^2 = 1+\frac {\pi}2.$$ 
A: What is $4-4+\cos^2x$? It is $\cos x$. 
$\cos^2x=\frac{cos 2x+1}{2}$
A: $$\sqrt{4-4 \sin^2\theta} = \sqrt{4 (1-\sin^2\theta)} = \sqrt{4 \cos^2\theta} = 2\cos\theta$$
Hence you get it as the above by using the substitution also.
Then use double angles formula to proceed. You shall get $2(\frac{1}{2}\sin 2\theta + \theta)$.
From the substitution, $x = \sin\theta$, you get $\cos\theta = \sqrt{1-x^2}$, hence $\sin 2\theta = x\sqrt{1-x^2}$.
Proceed further to get integral in terms of $x$.
