Jtdylktuy

I started with the substitution $x=sin(theta)$ with $thetain(-frac{pi}{2},frac{pi}{2})$

And I got the integral $$int_{-pi/2}^{pi/2} frac{1}{2-sin(theta)},dtheta$$
which I am again not able to evaluate.

Even integration by parts is not working.

Try using the substitution $displaystyle t = tan frac{theta}{2}$, this is a handy substitution to make when there are trigonometric functions that you cannot simplify very easily. In our case we have that,

$$frac{mathrm{d}t}{mathrm{d}theta} = frac{1}{2} sec^2frac{theta}{2} = frac{1}{2} left(1 + tan^2 frac{theta}{2} right) = frac{1}{2}(1 + t^2) Rightarrow frac{2mathrm{d}t}{1+t^2} = mathrm{d}theta $$

The usefulness of this substitution is that we can now use properties of a right angled triangle that has angle $theta /2$ to say that,

$$sin theta = 2sin frac{theta}{2} cos frac{theta}{2} = 2 frac{1}{sqrt{1+t^2}} frac{t}{sqrt{1+t^2}}= frac{2t}{1+t^2} $$

Putting this into our integral we have,

$$int_{-pi/2}^{pi/2} frac{mathrm{d}theta}{2 - sin theta} = int_{-1}^1 frac{2}{1+t^2} cdot frac{1}{2 - frac{2t}{1+t^2}} mathrm{d}t = int_{-1}^1 frac{1}{t^2 - t+ 1} mathrm{d}t $$

Now we have no more square roots, we have an irreducible quadratic at the bottom so we can complete the square to obtain,

$$int_{-1}^1 frac{1}{t^2 - t + 1} mathrm{d} t = int_{-1}^1 frac{1}{(t - frac{1}{2})^2 + frac{3}{4}} mathrm{d}t = frac{2}{sqrt{3}} tan^{-1} left(frac{2}{sqrt{3}}left(t - frac{1}{2}right) right) Big|_{-1}^1 = frac{pi}{sqrt{3}} $$

It is an integral of form $displaystyleint dfrac1{L sqrt{Q}}$, where L is a linear and Q is a quadratic. You could solve such integrals fairly easily with $L = 1/t$. Using this on your integral gives $displaystyle int dfrac{1}{sqrt{frac13 - (tsqrt 3 - frac2{sqrt{3}})^2}} dt$.

Not very appealing here but can be useful for other integrals of this type.

Here is a slightly different approach.

Let
$$I = int_{-1}^1 frac{dx}{(2 - x) sqrt{1 - x^2}} qquad (1)$$
Enforcing a substitution of $x mapsto - x$ gives
$$I = int_{-1}^1 frac{dx}{(2 + x) sqrt{1 - x^2}} qquad (2)$$
Adding (1) to (2) one obtains
$$I = 2 int_{-1}^1 frac{dx}{(4 - x^2) sqrt{1 - x^2}} = 4 int_0^1 frac{dx}{(4 - x^2) sqrt{1 - x^2}},$$
since the integrand is even between symmetric limits.

Now set $x = sin theta$. As $dx = cos theta , dtheta$, doing so yields
$$I = 4 int_0^{pi/2} frac{dtheta}{4 - sin^2 theta}.$$
Taking advantage of the identity $sin^2 theta + cos^2 theta = 1$, the dominator in the above integral can be rewritten as
$$I = 4 int_0^{pi/2} frac{dtheta}{3 sin^2 theta + 4 cos^2 theta} = 4 int_0^{pi/2} frac{sec^2 theta}{3 tan^2 theta + 4} , dtheta.$$
Setting $u = tan theta, du = sec^2 theta , dtheta$ one has
begin{align}
I &= 4 int_0^infty frac{du}{3u^2 + 4}\
&= frac{4}{3} int_0^infty frac{du}{u^2 + left (frac{2}{sqrt{3}} right )^2}\
&= frac{4}{3} left [frac{sqrt{3}}{2} tan^{-1} left (frac{u sqrt{3}}{2} right ) right ]_0^infty\
&= frac{pi}{sqrt{3}}.
end{align}

Here's a cool generalization consider the integral
$$J=I(x_1,x_2;a,b)=int_{x_1}^{x_2}frac{dx}{a+bsin x}$$
We may preform a tangent-half-angle substitution:
$$t=tan(x/2)Rightarrow dx=frac{2dt}{1+t^2}\Rightarrow sin x=frac{2t}{1+t^2}$$
Thus
$$J=2int_{t_1}^{t_2}frac{1}{a+2bfrac{t}{1+t^2}}frac{dt}{1+t^2}$$
Where $t_1=tan(x_1/2)$ and $t_2=tan(x_2/2)$. Anyway after a little algebra,
$$J=2int_{t_1}^{t_2}frac{dt}{at^2+2bt+a}$$
Then we complete the square in the denominator:
$$J=2int_{t_1}^{t_2}frac{dt}{a(t+frac{b}a)^2+a-frac{b^2}a}$$
then setting $g=a-frac{b^2}a$ and preforming the substitution
$$t+frac{b}a=sqrt{frac{g}a}tan uRightarrow dt=sqrt{frac{g}a}sec^2u, du$$
we see that
$$J=2sqrt{frac{g}a}int_{u_1}^{u_2}frac{sec^2u}{gtan^2u+g}du$$
Where $u_1=arctanbigg[sqrt{frac{a}g}bigg(t_1+frac{a}bbigg)bigg]$ and $u_2=arctanbigg[sqrt{frac{a}g}bigg(t_2+frac{a}bbigg)bigg]$. After we note that $frac{sec^2u}{gtan^2u+g}=frac1g$, we see that
$$J=frac2{sqrt{ag}}(u_2-u_1)$$
And after a bunch of algebra
$$I(x_1,x_2;a,b)=frac1{sqrt{a^2-b^2}}bigg[arctanfrac{atan(x_2/2)+b}{sqrt{a^2-b^2}}-arctanfrac{atan(x_1/2)+b}{sqrt{a^2-b^2}}bigg]$$
Which works as long as $a^2>b^2$.