Integrals Test - 18

$$\int_{0}^{1} tan  h x   dx$$
$$\int tan  h  x  dx=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}} dx =log (e^x+e^{-x})+c$$
$$\int_{0}^{1} tan  h x  dx = log (e^x+e^{-x})+c \int_{0}^{1}$$
$$=\left ( log (e+\dfrac{1}{e}) +c \right ) - \left ( log (2)+c \right )$$
$$=log (e+\dfrac{1}{e})-log 2$$
$$=log (\dfrac{e}{2}+\dfrac{1}{2e})$$
$$\int_{0}^{1}tan  hx  dx=log (\dfrac{e}{2}+\dfrac{1}{2e})$$

$$\int_{0}^{\pi}\dfrac{\tan  x}{\sec  x+\cos  x}dx$$

$$\int \dfrac{\tan  x}{\sec  x+\cos  x}dx=\int \dfrac{d  (\sec  x)}{1+\sec ^2 x}$$

$$=\tan ^{-1}(\sec  x)+c$$

$$\int_{0}^{\pi}\dfrac{\tan  x}{\sec  x+\cos  x}dx=\left [ \tan  (\sec  x)+c \right ] |_{0}^{\pi}$$

$$=(\tan ^{-1}(\sec  \pi)+c)- [ \tan ^{-1}(\sec  0) ]$$

$$=-\dfrac{\pi}{2}$$

Let D = $$\int_{0}^{1}e^x(e^x+1)^3dx$$
Consider $$I = \int e^x(e^x+1)^3dx=\int (e^x+1)^3e^x.dx=\frac{(e^x+1)^4}{4}+c$$
$$I = \int e^x(e^x+1)^3dx=\frac{(e^x+1)^4}{4}+c$$
$$D = \int_0^1 e^x(e^x+1)^3dx$$
    $$= \cfrac{(e^1+1)^4}{4} - 4$$
Thus, $$\int_{0}^{1}e^x(e^x+1)^3dx=\frac{(e+1)^4}{4}-4$$

$$\displaystyle\int_{\sqrt{8}}^{\sqrt{15}} x\sqrt{1+x^2} dx$$

$$\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}$$
$$\int \dfrac{x  dx}{(x^2+1)^2} =\dfrac{1}{2}\int \dfrac{dx^2}{(1+x^2)^2}=-\dfrac{1}{2}\left ( \dfrac{1}{1+x^2} \right )^{-1}$$
$$=-\dfrac{1}{2}(1+x^2)^{+c}$$
$$=\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}=\left [ -\dfrac{1}{2} (1+x^2)+c \right ] \int_{0}^{1}$$
$$=\left [ -\dfrac{1}{2} (2)+c \right ] - \left [ -\dfrac{1}{2}+c \right ]$$
$$=-\dfrac{1}{4}+\dfrac{1}{2} =\dfrac{1}{4}$$
$$\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}=\dfrac{1}{4}$$

$$ \displaystyle\int_{0}^{\dfrac{\pi}{2}} e^{\sin^2 x} \sin{2x} dx$$

$$\int_{0}^{4}\sqrt{16-x^2}  dx$$
$$\int \sqrt{16-x^2}  dx =\int \sqrt{4^2-x^2}dx$$
$$=\frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2} sin^{-1} (\frac{x}{14})+c$$
$$=\int_{0}^{4}\sqrt{16-x^2}dx  \left [ \frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2}sin^{-1}\left ( \frac{x}{4} \right ) +c \right ]\int_{0}^{4}$$
$$=\left ( \frac{4}{2}\sqrt{16-16}+ 8 sin^{-1}\left ( \frac{4}{4}\right ) +c\right )- (0+0+c)$$
$$=8\left ( \dfrac{\pi}{2} \right ) =4\pi$$
$$\int_{0}^{4}\sqrt{16-x^2}dx=4\pi$$

Let $$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \dfrac { \sec ^{ 2 }{ x }  }{ { \left( \sec { x+\tan { x }  }  \right)  }^{ n } }  } dx$$
Substitute $$\sec { x } +\tan { x } =t\Rightarrow \left( \sec { x\tan { x } +\sec ^{ 2 }{ x }  }  \right) dx=dt$$
$$\displaystyle \Rightarrow \sec { x } dx=\dfrac { dt }{ t } $$
$$\displaystyle \therefore \sec { x } -\tan { x } =\dfrac { 1 }{ t } \Rightarrow \sec { x } =\dfrac { 1 }{ 2 } \left( t+\dfrac { 1 }{ t }  \right) $$
$$\displaystyle \therefore I=\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty  }{ \dfrac { \left( t-\dfrac { 1 }{ t }  \right) \dfrac { dt }{ t }  }{ { t }^{ n } }  } =\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty  }{ \left( \dfrac { 1 }{ { t }^{ n } } +\dfrac { 1 }{ { t }^{ +n-12 } }  \right)  } dt$$

$$\displaystyle =\dfrac { 1 }{ 2 } \left[ \dfrac { 1 }{ \left( 1-n \right) { t }^{ n-1 } } +\dfrac { 1 }{ \left( n+1 \right) { t }^{ n+1 } }  \right] _{ 0 }^{ \infty  }=\dfrac { n }{ { n }^{ 2 }-1 } $$

$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4  cos^2 x  +  9  sec^2  x}$$
$$\int \dfrac{dx}{4  cos^2  x  + 9 sec^2  x}=\int \dfrac{sec^2 x  dx} {4+9  tan^2  x}=\int \dfrac{d(tan  x)}{4+9  tan^2  x}$$
$$=\dfrac{1}{4} tan^{-1}\left ( \dfrac{3}{2} tan  x \right )  \dfrac{2}{3} +c$$
$$=\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right )  +c$$
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4  cos^2 x  +  9  sec^2  x}=\left [ \dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan  x \right ) +c \right ]\int_{0}^{\dfrac{\pi}{2}}$$
$$=\left (\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan \dfrac{\pi}{2} \right ) +c \right ]- \left ( \dfrac{1}{6}tan^{-1}(0)+c \right )$$
$$=\dfrac{1}{6}\times \dfrac{\pi}{2}=\dfrac{\pi}{12}$$