Simple ?

Solution 1: Make the substitution $\sin x = \frac{2t}{1+t^{2}}$ where $t = \tan \frac{x}{2}$

Change the integration borders by substituting the values of x into t, which gives $\tan 0 = 0$ and $\tan \frac{\pi}{4} = 1$

Take the derivative of t with respect to x, giving $\frac{dt}{dx} = \frac{1}{2}\sec^{2}\frac{x}{2}$

Replacing with the identity $\sec ^{2} \theta = 1 + \tan ^{2} \theta$ we get $\frac{dt}{dx} = \frac{1 + t^{2}}{2}$

Rearranging for dx, we get $dx = \frac{2dt}{1+t^{2}}$

Returning back to the given integral, and substituting sinx, dx and the integration borders, we have $\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+\frac { 2t }{ 1+t^{2} } } } \frac { 2dt }{ 1+t^{2} }$

Simplifying, we get $\int _{ 0 }^{ 1 }{ \frac { 2dt }{ 1+t^{2}+2t } }$

Note that the denominator is a perfect square. Using negative indices, we get $\int _{ 0 }^{ 1 }{ 2(1+t)^{-2}dt }$

This can now be evaluated easily using the power rule, since the term being indexed is linear.

This yields $-2(1+t)^{ -1 }{ | }_{ 0 }^{ 1 }$

Converting back into a fraction and applying the borders, we get $\frac{-2}{1+1}-\frac{-2}{1+0}$

Which simplifies to $1$

Solution 2: Make the substitution $1+\sin \theta = 2\sin ^{2} \frac{2\theta + \pi}{4}$

Let $u = \frac{2x+\pi}{4}, du = \frac {dx}{2}$

Using the identity $\frac{1}{\sin\theta} = \csc\theta$ and all the substitutions above, the integral is now transformed into $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \csc^{2} u du }$

The integral of $\csc^{2} u du$ is $-\cot u$

The integral is now $- \cot u{ | }_{ 0 }^{ \frac{\pi}{2} }$

Substituting u back for x and evaluating the borders, we have $- \cot \frac{\pi + \pi}{4} + \cot \frac{\pi}{4}$

Which evaluates to $1$

Solution 3: Multiply the numerator and denominator by $1- \sin x$

Yielding $\frac{1- \sin x}{1-\sin^{2} x} = \frac{1- \sin x}{\cos^{2} x} = \sec^{2} x - \tan x \sec x$

The integral is now $\int _{ 0 }^{ \frac { \pi }{ 2 } }{\sec^{2} x - \sec x \tan x }$

Standard Integrals:

$\int {\sec x \tan x dx} = \sec x$

$\int {\sec^{2} x dx} = \tan x$

Resulting in: $(\tan x - \sec x) { | }_{ 0 }^{ \frac{\pi}{2} }$

Because we cannot evaluate this directly, we must transform the expression into something that can be evaluated.

$\tan x - \sec x = \frac{\tan^{2} x - \sec^{2} x}{\tan x + \sec x}$

Since $1 + \tan^2 x = \sec^2 x$ it follows that $\tan^{2} x - \sec^{2} x = -1$

$\frac{\tan^{2} x - \sec^{2} x}{\tan x + \sec x} = \frac{-1}{\tan x + \sec x} = \frac{-1}{\frac{\sin x +1}{\cos x}} = \frac{-\cos x}{1+ \sin x}$

This form can now be evaluated. Plugging in the borders:

$\frac{-\cos x}{1+ \sin x} { | }_{ 0 }^{ \frac{\pi}{2} } = \frac{0}{1+1} -(\frac{-1}{1+0}) = 1$