Integrals Test ...

The value of the integral $$\displaystyle \int_{0}^{\pi/4}\dfrac {\sin x + \cos x}{3 + \sin 2x}dx$$ is equal to

The value of $$\displaystyle \int_{0}^{\pi /2} \dfrac {\sin 2t}{\sin^{4}t + \cos^{4}t} dt $$

$$\displaystyle\int \dfrac{cos^{n-1}}{sin^{n+1}}dx, n\neq 0$$ is ____

$$\int { { ({ x }^{ 2 }+5) }^{ 3 } } dx$$

$$\displaystyle \int {\dfrac{dx}{\sin^2x \cos^2x}}$$

If $$\displaystyle \int_{\log 2}^{x} \dfrac {dy}{\sqrt {e^{y} - 1}} = \dfrac {\pi}{6}$$, then $$x$$ is equal to

Evaluate: $$\displaystyle \int \dfrac{(x-1)e^x}{(x+1)^3}dx=$$

Let $$f$$ be a positive function. Let
$${ I }_{ 1 }=\int _{ 1-k }^{ k }{ xf\left\{ x(1-x) \right\}  } dx$$
$${ I }_{ 2 }=\int _{ 1-k }^{ k }{ f\left\{ x(1-x) \right\}  } dx$$
where $$2k-1>0$$. Then $$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } $$

The value of $$\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 }  } }{ \dfrac { \sin ^{ -1 }{ x }  }{ { \left( 1-{ x }^{ 2 } \right)  }^{ { 3 }/{ 2 } } } dx } $$ is

Let $$f(x)$$ be a positive function. Let
$$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$$,
$$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$$,
where $$2k - 1 > 0$$, then $$\dfrac {I_{1}}{I_{2}}$$ is