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Integrals Test - 29

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Integrals Test - 29
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  • Question 1
    1 / -0
    The value of the integral $$\displaystyle \int_{0}^{\pi /4}\displaystyle \frac{\sin x+\cos x}{3+\sin 2x}dx$$ is
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ \pi /4 } \frac { \sin  x+\cos  x }{ 3+\sin  2x } dx=\int _{ 0 }^{ \pi /4 } \frac { \sin { x } +\cos { x }  }{ 4-{ \left( \sin { x } -\cos { x }  \right)  }^{ 2 } } dx$$

    Substitute $$\sin { x } -\cos { x } =t$$

    $$\displaystyle I=\int _{ -1 }^{ 0 }{ \frac { dt }{ 4-{ t }^{ 2 } }  } =\frac { 1 }{ 4 } \log { 3 } $$
  • Question 2
    1 / -0
    The value of $$\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}$$ is
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ \pi  } \frac { dx }{ 1-2\alpha \cos  x+\alpha ^{ 2 } } =\int _{ 0 }^{ \pi  } \frac { dx }{ 1-2\alpha \cos  x+\alpha ^{ 2 } } $$

    Substitute $$\displaystyle \tan { \frac { x }{ 2 }  } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 }  } dx=dt$$

    $$\displaystyle I=\int _{ 0 }^{ \infty  } \dfrac { 1 }{ 1-2\alpha \left( \dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }  \right) +\alpha ^{ 2 } } .\dfrac { 2t }{ 1+{ t }^{ 2 } } dt$$

    $$\displaystyle =\int _{ 0 }^{ \infty  } \dfrac { 2t }{ 1+{ t }^{ 2 }-2\alpha \left( 1-{ t }^{ 2 } \right) +\alpha ^{ 2 }\left( 1+{ t }^{ 2 } \right)  } dt$$

    $$\displaystyle =\frac { \pi  }{ { \alpha  }^{ 2 }-1 } $$ if $$\alpha >1$$
  • Question 3
    1 / -0
    If $$I= \displaystyle \int_{1}^{2}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}-1}}$$ then $$I$$ is equal to
    Solution
    As form of integrand is $$\displaystyle \dfrac { 1 }{ L\sqrt { Q }  } $$
    Substitute $$\displaystyle L=\dfrac { 1 }{ t } \Rightarrow x+1=\dfrac { 1 }{ t } $$

    $$\displaystyle I=\int _{ \dfrac { 1 }{ 2 }  }^{ \dfrac { 1 }{ 3 }  }{ \dfrac { \left( \dfrac { -1 }{ { t }^{ 2 } }  \right) dt }{ \dfrac { 1 }{ t } \sqrt { { \left( \dfrac { 1 }{ t } -1 \right)  }^{ 2 }-1 }  }  } =\int _{ \dfrac { 1 }{ 3 }  }^{ \dfrac { 1 }{ 2 }  }{ \dfrac { dt }{ \sqrt { 1-2t }  }  } $$

    $$\displaystyle ={ \left[ \dfrac { { \left( 1-2t \right)  }^{ \dfrac { 1 }{ 2 }  } }{ \left( -2 \right) \left( \dfrac { 1 }{ 2 }  \right)  }  \right]  }_{ \dfrac { 1 }{ 3 }  }^{ \dfrac { 1 }{ 2 }  }{ = }\dfrac { 1 }{ \sqrt { 3 }  } $$
  • Question 4
    1 / -0
    The value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{e^{x}+e^{-x}}$$ is
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dx }{ e^{ x }+e^{ -x } } =\int _{ 0 }^{ 1 } \frac { { e }^{ x }dx }{ e^{ 2x }+1 } $$

    Substitute $${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$$

    $$\displaystyle I=\int _{ 1 }^{ e }{ \frac { t }{ { t }^{ 2 }+1 } dt } ={ \left[ \tan ^{ -1 }{ t }  \right]  }_{ 1 }^{ e }=\tan ^{ -1 }{ e } -\frac { \pi  }{ 4 } $$
  • Question 5
    1 / -0
    The value of $$\displaystyle \int_{8}^{15}\displaystyle \frac{dx}{\left ( x-3 \right )\sqrt{x+1}}$$ is
    Solution
    $$\displaystyle I=\int _{ 8 }^{ 15 } \frac { dx }{ \left( x-3 \right) \sqrt { x+1 }  } $$

    Substitute $$\sqrt { x+1 } =t\Rightarrow x+1={ t }^{ 2 }$$

    $$\displaystyle I=\int _{ 3 }^{ 4 }{ \frac { 2t }{ \left( { t }^{ 2 }-4 \right) t } dt } ={ \left[ \frac { 2 }{ 2.2 } \log { \left( \frac { t-2 }{ t+2 }  \right)  }  \right]  }_{ 3 }^{ 4 }$$

    $$\displaystyle =\frac { 1 }{ 2 } \left[ \log { \frac { 1 }{ 3 }  } -\log { \frac { 1 }{ 5 }  }  \right] =\frac { 1 }{ 2 } \log { \frac { 5 }{ 3 }  } $$
  • Question 6
    1 / -0
    If $$b > a,$$ and $$\displaystyle  I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,$$ then $$I$$ equals
    Solution
    Substitute $$ b -x = t^{2} $$ so that
    $$\displaystyle I = \int _{\sqrt{b-a}}^{0}\sqrt{\frac{b-t^{2}-a}{t^{2}}} (-2t) dt$$
    $$ \displaystyle= 2 \int_{0}^{c} \sqrt{c^{2}-t^{2}} dt $$,  where $$ c= \sqrt{b-a}$$
    $$=\displaystyle

    2 \left [ \frac{1}{2} t \sqrt{c^{2}-t^{2}}+ \frac{c^{2}}{2} \sin ^{-1}

    \left( \frac{t}{c}\right) \right] _{0}^{c}$$
    $$ =0 + c^{2} \sin^{-1} (1) -0$$ $$\displaystyle =\frac{\pi}{2}(b-a) $$
  • Question 7
    1 / -0
    f $$ k = e^{2007} $$ then value of $$ \displaystyle I =\int_{1}^{k}\frac{ \pi \cos (\pi \log x )} {x} dx $$ is
    Solution
    $$ \displaystyle I =\int_{1}^{k}\frac{ \pi \cos (\pi \log x )} {x} dx $$
    Substitute $$\pi \log x = \theta \Rightarrow \cfrac{\pi}{x}dx = d\theta$$
    $$ \therefore I=\displaystyle  \int_{0}^{2007} \cos \theta d \theta = \sin \theta ] _{0}^{2007\pi}=0 $$
  • Question 8
    1 / -0
    Value of $$\displaystyle \int_{0}^{25}\displaystyle \frac{1}{\sqrt{4+\sqrt{x}}}\: dx$$ is
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ 25 }{ \frac { 1 }{ \sqrt { \sqrt { x } +4 }  } dx } $$

    Substitute $$\displaystyle  u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x }  } dx$$

    $$\displaystyle  I=2\int _{ 0 }^{ 5 }{ \frac { u }{ \sqrt { u+4 }  } du } $$

    Substitute $$ s=u+4\Rightarrow ds=du$$

    $$\displaystyle I=2\int _{ 4 }^{ 9 }{ \frac { s-4 }{ \sqrt { s }  } ds } =2\int _{ 4 }^{ 9 }{ \left( \sqrt { s } -\frac { 4 }{ \sqrt { s }  }  \right) ds } $$

    $$\displaystyle={ \left[ \frac { 4{ s }^{ 3/2 } }{ 3 }  \right]  }_{ 4 }^{ 9 }-{ \left[ 16\sqrt { s }  \right]  }_{ 4 }^{ 9 }=36-\frac { 32 }{ 3 } -48-32=-\frac { 164 }{ 3 } $$
  • Question 9
    1 / -0
    $$\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$$
    Solution
    $$I = \displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx $$

    Substitute $$\displaystyle x = \frac{1}{t}\Rightarrow dx =-\frac{1}{t^2}dt$$

    $$I = \displaystyle \int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln (1/t)}{1/t} .\frac{-dt}{t^2} $$

    $$\quad \displaystyle =\int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln t}{t} dt$$

    $$\quad \displaystyle =\int_{\infty }^0 f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x} dx=-I$$

    $$\Rightarrow 2I = 0\Rightarrow I = 0$$
  • Question 10
    1 / -0
    If$$\displaystyle \int_{0}^{\pi /3}\frac{\cos }{3+4\sin x}dx=K\log \frac{\left ( 3+2\sqrt{3} \right )}{3}$$ then K is
    Solution
    $$\displaystyle I=\int _{ 0 }^{ \pi /3 }{ \cfrac { \cos { x }  }{ 3+4\sin { x }  } dx } $$

    Substituting $$t=3+4\sin { x } \Rightarrow dt=4\cos { x } $$

    $$\displaystyle I=\frac { 1 }{ 4 } \int _{ 3 }^{ 3+2\sqrt { 3 }  }{ \frac { 1 }{ t } dt } ={ \left[ \frac { \log { t }  }{ 4 }  \right]  }_{ 3 }^{ 3+2\sqrt { 3 }  }$$

    $$\displaystyle=\frac { \log { \left( 3+2\sqrt { 3 }  \right)  } -\log { 3 }  }{ 4 } =\frac { 1 }{ 4 } \log { \left( \frac { 3+2\sqrt { 3 }  }{ 3 }  \right)  } \Rightarrow K=\frac { 1 }{ 4 } $$
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