Definite integral $\int_{0}^{\pi/2}\ 1/ (1+(\tan x)^{1/2})\ dx$ $$\int_{0}^{\pi/2}\ \frac{1}{ 1+(\tan x)^{1/2}}\ dx$$
I have no idea how to evaluate this. I have tried many substitutions, but they just didn’t result in the answer.
Update:
As a remainder, if one wants to integrate a similar question, $\int_{0}^{\pi/2}\ 1/ (1+(tanx)^{\sqrt2})\ dx$ , refer to this  Evaluate $\int_0^\pi\frac{1}{1+(\tan x)^\sqrt2}\ dx$ .
Both of them were what I want to ask. I have tried many ways to find their anti-derivatives; however, in cases of this type (definite integral with a complicated integrand), their indefinite integral could not even be expressed in basic functions, let alone use Fundamental theorem of calculus II to evaluate them. Suitably using the brilliant method given below can directly lead to the answer.
Thank you very much.
 A: There is a trick you can use here.
Let $I$ be the integral you want. It turns out that $\int_0^\frac{\pi}{2} \frac{\mathrm{d}x}{1 + \mathrm{cot}(x)^{0.5}}$ is also equal to $I$. You can see this by substituting $y = \frac{\pi}{2} - x$ in the original equation, or by drawing graphs of the tangent and cotangent functions and seeing that they are symmetric in this domain - both methods are essentially the same.
So we can add the two.
$$
2I = \int_0^\frac{\pi}{2} \left(\frac{1}{1 + \mathrm{tan}(x)^{0.5}} + \frac{1}{1 + \mathrm{cot}(x)^{0.5}} \right) \mathrm{d}x
$$
Since $\mathrm{tan}(x) = \mathrm{cot}(x)^{-1}$, this simplifies to 
$$
2I = \int_0^\frac{\pi}{2} \left(\frac{1}{1 + \mathrm{tan}(x)^{0.5}} + \frac{\mathrm{tan}(x)^{0.5}}{\mathrm{tan}(x)^{0.5} +1} \right) \mathrm{d}x
$$
Which works out (everything cancels!) to $2I = \int_0^\frac{\pi}{2} \mathrm{d}x$
So $I = \frac{\pi}{4}$. Funnily enough, this is true whatever be the power to which the tangent is raised.
A: Try to write $\sqrt {tanx} $ in terms of sin x and cos x i.e $$\sqrt {tanx} = {\sqrt {sinx} \over \sqrt {cosx}}$$
Then multiply numerator and denominator of integrand with $\sqrt {cosx}$ 
Then integral turns 
$$\int_0^{\pi \over 2} {\sqrt {cosx} \over \sqrt {sinx}+\sqrt {cosx}} dx=I..(let)$$
Now use kings rule $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
Then the integrand becomes$$\int_0^{\pi \over 2} {\sqrt {sinx} \over \sqrt {sinx}+\sqrt {cosx}}dx=I $$
Add these two forms of $I$
Now you have$$2I=\int_0^{\pi \over 2} 1 dx$$
Which is pretty easy to evaluate so you get$$I={\pi \over 4}$$
Thats the answer 
$$\int_{0}^{\pi/2}\ 1/ (1+(tanx)^{1/2})\ dx= {\pi \over 4}$$
You may notice that the exponent of tan x is irrelevant here as it does not affect any of the above calculation
A: Let $u = \tan x$, then $\mathrm{d}u = \sec^2 x~\mathrm{d}x$ and $\mathrm{d}x = \dfrac{\mathrm{d}u}{1+u^2}$. So:
$$I=\int_{0}^{\pi/2}\frac{1}{1+\sqrt{\tan x}}\mathrm{d}x=\int_{0}^{\infty}\frac{1}{(1+\sqrt{u})(1+u^2)}\mathrm{d}u$$
then let $t=\sqrt{u}$ where $\mathrm{d}t=\dfrac{1}{2\sqrt{u}}\mathrm{d}u$ and the integral becomes
$$I=\int_0^{\infty}\frac{2t}{(1+t)(1+t^4)}\mathrm{d}t$$
which can then be solved by repeated partial fraction decomposition. Alternatively, a similar approach taken to the one in Hrishabh's answer seems to work out much better:
$$I=\int_{0}^{\pi/2}\frac{1}{1+\sqrt{\tan x}}\mathrm{d}x=\frac{1}{1+\sqrt{\frac{\sin x}{\cos x}}}\mathrm{d}x$$
multiply $I$ by $\sqrt{\cos x}$ to form
$$I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\mathrm{d}x\tag{1}$$
then apply
$$\int_a^b f(x)\mathrm{d}x=\int_a^b f(a+b-x)\mathrm{d}x$$
to form
$$I=\int_{0}^{\pi/2}\frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}\mathrm{d}x$$
where since $\cos \left(\frac{\pi}{2}-x\right)=\sin x$ and $\sin\left(\frac{\pi}{2}-x\right)=\cos x$ we get
$$I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\mathrm{d}x\tag{2}$$
therefore adding $(1)$ and $(2)$
$$2I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\mathrm{d}x=\int_{0}^{\pi/2}\mathrm{d}x$$
hence
$$I=\frac{\pi}{4}$$
which certainly appears to be a better approach than dealing with the partial fraction decomposition in the first method.
