Integrals Test ...

Suppose that F(x) is an antiderivative of f(x)$$\displaystyle =\frac{\sin x}{x},x> 0$$ then $$\displaystyle \int_{1}^{3}\frac{\sin 2x}{x}$$ can be expressed as

If $$0 < \alpha < 1$$ and $$\displaystyle I=\int _{-1}^{1} \frac{dx}{\sqrt{1-2\alpha x+\alpha^{2}}} $$ then $$I$$ equals

$$\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx$$ has the value equal to 

The value of the definite integral $$\displaystyle \int_{1}^{\infty}\left ( e^{x+1}+e^{3-x} \right )^{-1}dx$$ is

$$\displaystyle \int ^{\displaystyle \frac{3 \pi}{10}}_{\displaystyle \frac{\pi}{5}} \frac{sin x}{sin x + cos x}dx $$ is equal to 

$$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}$$ equals

$$I_{1}$$ is equal to

Evaluate $$\displaystyle \int_{0}^{\pi /2} \frac{dx}{2+\sin 2x}$$

$$\displaystyle \int_{\tfrac{1}{\sqrt{3}}}^{0}\dfrac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$

$$\int \dfrac {1}{x^{2} (x^{4} + 1)^{3/4}}dx$$ is equal to