Integrals Test - 21

$$\displaystyle \int_{\frac {1}{3}}^{1} \dfrac {(x – x^{3})^{\frac {1}{3}}}{x^{4}} dx = \int_{\frac {1}{3}}^{1} \frac {(x^{3})^{\frac {1}{3}} \left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{4}} dx$$

$$\displaystyle = \int_{\dfrac {1}{3}}^{1} \dfrac {\left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{3}} dx$$           (let $$\dfrac {1}{x^{2}} – 1 = t$$)

$$\displaystyle = \int_{8}^{0} \dfrac{t^{\frac{1}{3}}}{-2} dt$$    $$\therefore \dfrac {-2}{x^{3}} dx = dt$$

$$= -\dfrac {1}{2}\left (\dfrac {t^{4/3}}{4/3}\right )_{8}^{0} = 6$$.

Let $$I=\int { \cfrac { dx }{ \sqrt { { e }^{ x }-1 }  }  } $$
substitute $$u={ e }^{ x }\Rightarrow du={ e }^{ x }dx$$, we get
$$I=\int { \cfrac { 1 }{ u\sqrt { u-1 }  } du } $$
Substitute $$s=u-1\Rightarrow ds=du$$, we get
$$I=\int { \cfrac { ds }{ \sqrt { s } \left( s+1 \right)  }  } $$
Substitute $$p=\sqrt { s } \Rightarrow dp=\cfrac { 1 }{ 2\sqrt { s }  } ds$$, we get
$$I=2\int { \cfrac { 1 }{ { p }^{ 2 }+1 }  } =2{ tan }^{ -1 }p=2{ tan }^{ -1 }\sqrt { s } =2{ tan }^{ -1 }\sqrt { u-1 } $$
$$=2{ tan }^{ -1 }\sqrt { { e }^{ x }-1 } $$
Therefore,
$$\int _{ \log { 2 }  }^{ t }{ \cfrac { dx }{ \sqrt { { e }^{ x }-1 }  }  } =\cfrac { \pi  }{ 6 } \\ \Rightarrow { \left[ 2{ tan }^{ -1 }\sqrt { { e }^{ x }-1 }  \right]  }_{ \log { 2 }  }^{ t }=\cfrac { \pi  }{ 6 } \Rightarrow 2{ tan }^{ -1 }\sqrt { { e }^{ t }-1 } -2{ tan }^{ -1 }1=\cfrac { \pi  }{ 6 } \\ \Rightarrow 2{ tan }^{ -1 }\sqrt { { e }^{ t }-1 } -\cfrac { \pi  }{ 2 } =\cfrac { \pi  }{ 6 } \Rightarrow { tan }^{ -1 }\sqrt { { e }^{ t }-1 } =\cfrac { \pi  }{ 3 } \\ \Rightarrow { e }^{ t }-1=3\Rightarrow t=\log { 4 } $$

Let, $$x=\cos\theta\implies dx=-sin\theta d\theta$$

$$\displaystyle\int_{0}^{\pi/2} \sin^3{x}.\cos^3{x} dx$$

$$\int_{0}^{\infty}\left ( \dfrac{a^x}{c^x}-\dfrac{b^x}{c^x} \right ) dx$$
$$=\int_{0}^{\infty}\left ( \dfrac{a}{c} \right )^xdx-\int_{0}^{\infty}\left ( \dfrac{b}{c}\right )^x dx$$
$$=\dfrac{(\dfrac{a}{c})^x}{log (\dfrac{a}{c})}\int_{0}^{\infty}-\dfrac{(\dfrac{b}{c})^x}{log (\dfrac{b}{c})}\int_{0}^{\infty}$$
$$=\left [ 1-\dfrac{1}{log(\dfrac{a}{c})} \right ] - \left [ 1-\dfrac{1}{log(\dfrac{b}{c})}\right ]$$
$$=\dfrac{1}{log(\dfrac{b}{c})}-\dfrac{1}{log(\dfrac{a}{c})}$$

$$\displaystyle \int_{0}^{1}e^x\left ( \dfrac{1}{(x=+1)}-\dfrac{1}{(x+1)^2}\right ) dx$$

It is in the form of

$$\displaystyle \int  e^x \left [ f(x)+f^1 (x) \right ]  dx=e^x f(x)+c$$

$$=\left [ \dfrac{e^x}{(1+x)}+c \right ] _{0}^{1}$$

$$=\dfrac{e}{2}-1$$

$$=\dfrac{e}{2}-1$$

$$\int_{\sqrt{2}}^{x}\dfrac{dx}{x\sqrt{x^2-1}}=\dfrac{\pi}{2}$$
$$=\left ( sin^{-1}x+c \right ) \int_{\sqrt{2}}^{x}=\dfrac{\pi}{2}$$
$$\left ( sec^{-1}x - sec^{-1}\sqrt{2} \right ) =\dfrac{\pi}{2}$$
$$sec^{-1}x- \dfrac{\pi}{4}=\dfrac{\pi}{12}$$
$$sec^{-1}x =\dfrac{\pi}{12}+\dfrac{3\pi}{12}$$
$$sec^{-1}=\dfrac{\pi}{3}$$
$$x=sec  \dfrac{\pi}{3}$$
$$=2$$

$$\int_{0}^{1}\dfrac{dx}{x^2+x  cos_\alpha+1}$$
$$\int_{0}^{1}\dfrac{dx}{(x+cos  \alpha)^2}+ sin^2    \alpha$$
$$\left ( \dfrac{1}{sin  \alpha} \left [ tan^{-1} \left ( \dfrac{x+ cos  \alpha}{sin  \alpha} \right ) \right ]+c \right )_{0}^{1}$$
$$=\dfrac{1}{sin  \alpha} \left [ tan^{-1} \left ( \dfrac{1+cos  \alpha}{sin  \alpha} \right )  \right ]-\dfrac{1}{sin  \alpha}tan^{-1}\left ( \dfrac{cos  \alpha}{sin  \alpha} \right )$$
$$=\dfrac{1}{sin  \alpha}\left [ tan^{-1} (cot  \dfrac{\alpha}{2})-tan^{-1}(cot  \alpha) \right ]$$
$$=\dfrac{1}{sin \alpha} \left ( \alpha -\dfrac{\alpha}{2} \right ) =\dfrac{\alpha}{2} sin  \alpha$$

$$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  sin^2  x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  cos^2  x}dx$$
$$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  cos^2x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{sec^2x}{(sec^2x + 4)}dx$$
$$=\int_{0}^{\dfrac{\pi}{4}} \dfrac{d(tan   x)}{5 + tan^2  x} dx= \left [ \dfrac{1}{\sqrt{5}} tan^{-1}\left ( \dfrac{tan  x}{\sqrt{5}} \right ) + c \right ]_{0}^{\dfrac{\pi}{2}}$$
$$=\dfrac{1}{\sqrt{5}}(\dfrac{\pi}{2})$$
$$=\dfrac{\pi}{2\sqrt{5}}$$