Integrals Test - 22

$$\int_{0}^{16}\dfrac{dx}{\sqrt{x+9}-\sqrt{x}}$$
$$\int_{0}^{16}\dfrac{\sqrt{x}+9+\sqrt{x}}{9}dx$$
$$=\dfrac{1}{19}\left [ \int_{0}^{16} \sqrt{x+9}dx + \int_{0}^{16}\sqrt{x}dx \right ]$$
$$=\dfrac{1}{19}\times \dfrac{2}{3}\left [ (x+a)^{\dfrac{3}{2}}\int_{0}^{16}+ x^{\dfrac{3}{2}} \int_{0}^{16} \right ]$$
$$=\dfrac{2}{27}\left [ (25)^{\dfrac{3}{2}} - (9)^{\dfrac{3}{2}}+ (16)^{\dfrac{3}{2}} \right ]$$
$$=\dfrac{2}{27}\left [ 125-27+64 \right ]$$
$$=\dfrac{2}{27}\left [ 189-27 \right ] =\dfrac{2}{27}\left [ 162 \right ]$$
$$=12$$

Let $$I=\int { \cfrac { \sqrt { x }  }{ 1+x } dx } $$
Substitute $$t=\sqrt { x } \Rightarrow dt=\cfrac { 1 }{ 2\sqrt { x }  } ds$$
$$I=2\int { \cfrac { { t }^{ 2 } }{ { t }^{ 2 }+1 } dt } =2\int { \left( 1-\cfrac { 1 }{ { t }^{ 2 }+1 }  \right) dt } =2\int { dt } -2\int { \cfrac { 1 }{ { t }^{ 2 }+1 } dt } \\ =2t-2{ tan }^{ -1 }t=2\sqrt { x } -2{ tan }^{ -1 }\sqrt { x } $$
Therefore
$$\int _{ 0 }^{ 1 }{ Idx } ={ \left[ 2\sqrt { x } -2{ tan }^{ -1 }\sqrt { x }  \right]  }_{ 0 }^{ 1 }=2-\cfrac { \pi  }{ 2 } $$

$$\int_{0}^{\dfrac{\pi}{4}}\dfrac{sec^2  x \sqrt{tan  x}}{tan  x}dx$$
dividing  and multiplying by $$sec^2 x$$
$$=\int_{0}^{\dfrac{\pi}{4}}\dfrac{d  tan  x}{\sqrt{tan  x}}$$
$$=(2\sqrt{tanx}+c)\int_{0}^{\dfrac{\pi}{4}}$$
$$=2$$

$$\int_{1}^{2}\dfrac{dx}{(x-1)^2+(\sqrt{3})^2}=\left [ \dfrac{1}{\sqrt{3}} tan^{-1} \left ( \dfrac{x-1}{\sqrt{3}}\right ) +c \right ] \int_{1}^{2}$$
$$=\dfrac{1}{\sqrt{3}}tan^{-1}\left ( \dfrac{1}{\sqrt{3}} \right )$$
$$=\dfrac{1}{\sqrt{3}}\cdot \dfrac{\pi}{6}$$
$$=\dfrac{\pi}{6\sqrt{3}}$$

$$x=t^2$$
$$dx=2t   dt$$
$$=\int_{0}^{1}\left ( \dfrac{t^2}{1+t} \right )  2t \cdot  dt$$
$$=2\int_{0}^{1}\dfrac{t^3}{1+t}dt =2\int_{0}^{1}\dfrac{t^3+1}{t+1}dt -2\int_{0}^{1}\dfrac{1}{t+1}dt$$
$$=2\left [ \dfrac{t^3}{3}\dfrac{t^2}{2}+t \right] \int_{0}^{1} -2 log (1+1) \int_{0}^{1}$$
$$=2\left [ \dfrac{1}{3} -\dfrac{1}{2} +1 \right ] -2 log (2)$$
$$=2 \left [ \dfrac{2-3+6}{6} \right ] -2  log (2)$$
$$=\dfrac{5}{3}-log  4$$

$$\int_{0}^{\pi}\dfrac{dx}{3+2  sin  x+  cos  x}$$
$$=\int_{0}^{\pi}\dfrac{dx(1+tan^2 \dfrac{x}{2})}{3(1+tan^2 \dfrac{x}{2}) + 2 (2 tan \dfrac{x}{2}) + (1- tan^2 \dfrac{x}{2})}$$
$$=\int_{0}^{\pi}\dfrac{sec^2\dfrac{x}{2} dx}{2 tan^2 \dfrac{x}{2}+ 4 tan \dfrac{x}{2}+4}$$
$$=\int_{0}^{\pi}\dfrac{2 d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d(tan^2 \dfrac{x}{2})}{(tan \dfrac{x}{2}+1)^2} =tna^{-1}(tan \dfrac{x}{2}+1)\int_{0}^{\pi}$$
$$=\dfrac{\pi}{2} -\dfrac{\pi}{4}$$
$$=\dfrac{\pi}{4}$$

Consider, $$I=\displaystyle \int_{0}^{1}\sin \left (2 tan^{-1}\sqrt{\dfrac{1-x}{1+x}} \right ) dx$$

let, $$x=  \cos \theta$$ $$\Rightarrow$$ $$dx=-\sin \theta  d \theta$$

$$\Rightarrow$$ $$x\rightarrow 0 \Rightarrow  \cos \theta \rightarrow \dfrac{\pi}{2}$$ and $$x\rightarrow 1 \Rightarrow  \cos \theta \rightarrow 0$$

$$\Rightarrow$$ $$I=\displaystyle \int_{\frac{\pi}{2}}^{0} \sin ( 2 tan^{-1}(tan(\dfrac{\theta}{2})) (-\sin \theta)  d\theta$$

$$\Rightarrow$$ $$I=\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^2 \theta  d \theta = \displaystyle \int_{0}^{\frac{\pi}{2}}\left [ \dfrac{1-\cos 2 \theta }{2} \right ] d\theta$$

$$\Rightarrow$$ $$I=\displaystyle \int_{0}^{\frac{\pi}{2}}\dfrac{1}{2}d  \theta - \dfrac{1}{2}\displaystyle \int_{0}^{\frac{\pi}{2}}\cos 2 \theta d \theta$$

$$\Rightarrow$$ $$I=\left[\left ( \dfrac{\pi}{4}-0 \right )+\dfrac{1}{2}  \sin 2 \theta \right]_{0}^{\frac{\pi}{2}}$$

$$\Rightarrow$$ $$I=\dfrac{\pi}{4}$$

$$\displaystyle\int_{0}^{\pi/4} \dfrac{\sin^{\frac{1}{2}}x}{\cos^{\frac{5}{2}}x} dx$$

$$\int_{0}^{\dfrac{\pi}{2}}\left ( 2  tan \dfrac{x}{2} + x  sec^2  \dfrac{x}{2} \right )  dx$$
$$=2 \int_{0}^{\dfrac{\pi}{2}}\left ( tan \dfrac{x}{2} + \dfrac{1}{2} x sec^2  \dfrac{x}{2} \right )  dx$$
$$=2 \int_{0}^{\dfrac{\pi}{2}}d\left ( x  tan \dfrac{x}{2} \right )$$
$$=2  \left ( x  tan \dfrac{x}{2} \right ) \int_{0}^{\dfrac{\pi}{2}}$$
$$=2 \left [ \left ( \dfrac{\pi}{2} 1 \right ) - 0 \right ]$$
$$=\pi$$

Let $$I=\int { \cfrac { 2-x }{ \sqrt { 5x-6-{ x }^{ 2 } }  } dx } =\int { \cfrac { 2-x }{ \sqrt { \cfrac { 1 }{ 4 } -\left( x-\cfrac { 5 }{ 2 }  \right) ^{ 2 } }  } dx } $$
Substitute $$t=x-\cfrac { 5 }{ 2 } \Rightarrow dt=dx$$
$$I=\int { \cfrac { -t-\cfrac { 1 }{ 2 }  }{ \sqrt { \cfrac { 1 }{ 4 } -{ t }^{ 2 } }  } dt } $$
Substitute $$t=\cfrac { sinu }{ 2 } \Rightarrow dt=\cfrac { cosu }{ 2 }du $$
$$I=\int { \left( -\cfrac { sinu }{ 2 } -\cfrac { 1 }{ 2 }  \right) du } =\cfrac { cosu }{ 2 } -\cfrac { u }{ 2 } =\cfrac { cos\left( { sin }^{ -1 }2t \right)  }{ 2 } -\cfrac { { sin }^{ -1 }2t }{ 2 } \\ =\cfrac { cos\left( { sin }^{ -1 }2\left( x-\cfrac { 5 }{ 2 }  \right)  \right)  }{ 2 } -\cfrac { { sin }^{ -1 }2\left( x-\cfrac { 5 }{ 2 }  \right)  }{ 2 } =\cfrac { cos\left( { sin }^{ -1 }\left( 2x-5 \right)  \right)  }{ 2 } -\cfrac { { sin }^{ -1 }\left( 2x-5 \right)  }{ 2 } $$
Therefore
$$\int _{ 2 }^{ 3 }{ Idx= } { \left[ \cfrac { cos\left( { sin }^{ -1 }\left( 2x-5 \right)  \right)  }{ 2 } -\cfrac { { sin }^{ -1 }\left( 2x-5 \right)  }{ 2 }  \right]  }_{ 2 }^{ 3 }=0-\cfrac { \pi  }{ 4 } -0-\cfrac { \pi  }{ 4 } =-\cfrac { \pi  }{ 2 } $$