Integrals Test ...

$$\displaystyle \int_{\pi /4}^{3\pi /4}\frac{dx}{1+\cos x}$$ is equal to

If $$I_1 = \displaystyle \int_x^1 \frac{1}{1 + t^2} dt$$ and $$I_2 \displaystyle = \int_1^{1 / x}\frac{1}{1+ t^2}dt$$ for $$x > 0$$, then

If $$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$$ for $$x> 0,$$ 

$$ \displaystyle \int_{\sin \theta }^{\cos \theta }f(x \tan \theta )dx\left ( where \theta \neq \frac{n\pi}{2} ,n\epsilon I\right )$$ is equal to

Suppose that F(x) is an anti-derivative of $$\displaystyle f(x)=\frac{\sin x}{x}$$, where $$x>0$$.

If $$\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$$ then $$\displaystyle {f}'\left ( \frac{\pi }{3} \right )$$ is

The solution for x of the equation $$\displaystyle \int_{\sqrt{2}}^{x}\frac{dt}{t\sqrt{t^{2}-1}}=\frac{\pi }{2}$$ is

$$\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx$$

$$\displaystyle \int_{1}^{2}\left ( x+\frac{1}{x} \right )^{3/2}\frac{x^{2}-1}{x^{2}}dx$$

The value of$$\displaystyle \int_{1}^{2}\frac{\cos \left ( \log x \right )}{x}dx$$  is equal to