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Integrals Test - 54

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Integrals Test - 54
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  • Question 1
    1 / -0
    If $$\displaystyle  I =\int_{0}^{a} \sqrt {\frac{a-x}{a+x}}dx, a > 0,$$ then $$I$$ equals
    Solution
    $$\displaystyle  I =\int_{0}^{a} \sqrt {\frac{a-x}{a+x}}dx$$

    $$I=\displaystyle \int _{ 0 }^{ a } \sqrt { \frac { a-x }{ a+x } \times \frac { a-x }{ a-x }  } dx$$

    $$I=\displaystyle \int _{ 0 }^{ a } \frac { a-x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } }  } dx$$

    $$I=\displaystyle \int _{ 0 }^{ a } \frac { a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } }  } dx-\int _{ 0 }^{ a } \frac { x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } }  } dx$$

    $$I=a{ [\sin ^{ -1 }{ \dfrac { x }{ a } ] }  }_{ 0 }^{ a }-\int _{ 0 }^{ a } \dfrac { x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } }  } dx$$

    Substitute $${ a }^{ 2 }-{ x }^{ 2 }=t^2$$
    $$\Rightarrow -xdx=tdt$$
    When $$x=0 \Rightarrow t=a$$
    When $$x=a \Rightarrow t=0$$

    $$I=\displaystyle a\left[\sin ^{ -1 }{ \dfrac { x }{ a }} \right]_{ 0 }^{ a }-\int _{ 0 }^{ a } \dfrac { t }{ \sqrt { { t }^{ 2 } }  } dt$$

    $$I=a\dfrac { \pi  }{ 2 } -a$$
    $$\Rightarrow I=a\left(\dfrac { \pi  }{ 2 } -1 \right)$$

  • Question 2
    1 / -0
    Value of $$\displaystyle \int_{0}^{\pi /2}\displaystyle \frac{\sin 4\Theta }{\sin \Theta }\: d\Theta $$ is
    Solution
    $$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \sin { 4\Theta  }  }{ \sin { \Theta  }  } d\Theta  } =\int _{ 0 }^{ \pi /2 }{ \sin { 4\Theta  } \csc { \Theta  } d\Theta  } $$

    $$=\int _{ 0 }^{ \pi /2 }{ \left( 2\cos ^{ 3 }{ \Theta  } +2\cos { \Theta  } -6\sin ^{ 2 }{ \Theta  } \cos { \Theta  }  \right) d\Theta  } $$

    $$\displaystyle ={ \left[ -2\sin ^{ 3 }{ \Theta  } +\frac { 10\sin { \Theta  }  }{ 3 } +\frac { 2 }{ 3 } \sin { \Theta  } \cos ^{ 2 }{ \Theta  }  \right]  }_{ 0 }^{ \pi /2 }=\frac { 4 }{ 3 } $$
  • Question 3
    1 / -0
    If $$ \displaystyle I = \int_{0}^{\pi/2} \frac{dx}{5+3\sin x}  =\lambda \tan^{-1} \left(\frac{1}{2}\right ) $$ then
    value of $$\lambda $$ is
    Solution
    $$ \displaystyle I = \int_{0}^{\pi/2} \frac{dx}{5+3\cos x} $$

    Substitute $$ \tan (x/2)= t,$$ so that

    $$\displaystyle dx=\frac{2\:dt}{1+t^{2}}$$ and $$\displaystyle 5 +3 \cos x= \frac{8+2t^{2}}{1+t^{2}} $$

    $$ \displaystyle \therefore I = \int_{0}^{1} \frac{dt}{4+t^{2}} =\frac{1}{2}\tan^{-1}\left. \left(\frac{t}{2}\right)\right ]_{0}^{1}$$

    $$\displaystyle =\frac{1}{2} \tan^{-1} \left( \frac{1}{2}\right) $$ 

    Thus, $$ \lambda =1/2$$
  • Question 4
    1 / -0
     If $$\displaystyle I = \int_{1/3}^{3}\frac{1}{x}\sin \left (\frac{1}{x}-x \right) dx,$$ then $$I$$ equals
    Solution
    Substitute $$ x =1/t$$ so that
    $$\displaystyle  I= \int_{3}^{1/3}  t \sin \left (t-\frac{1}{t} \right) \frac{(-1)}{t^{2}}  dt$$
    $$ \displaystyle = \int_{1/3}^{3}  \frac{1}{t}  \sin \left (\frac{1}{t} -t\right) dt= -I $$
    $$\Rightarrow 2I = 0 $$ or $$ I =0$$
  • Question 5
    1 / -0
     If $$\displaystyle I=\int _{8}^{15} \frac{dx}{(x-3)\sqrt{x+1} }$$ then$$ I$$ equals
    Solution
    Put $$\sqrt{x+1} =t $$ or $$ x + 1 =t^{2} $$
    $$\displaystyle

    \therefore I=\int _{3}^{4} \frac{2t}{(t^{2}-4) t}

    dt=\frac{2}{(2)(2)}\log \left. \left | \frac{t-2}{t+2}\right |\right

    ]_{3}^{4}$$
    $$\displaystyle =\frac{1}{2} \left[ \log \frac{1}{3}-\log \frac{1}{5}\right] $$
    $$\displaystyle =\frac{1}{2}\log \frac{5}{3}$$
  • Question 6
    1 / -0
    37 If $$ n > 1,$$ and $$ \displaystyle I=\int _{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^{2}})^{n}}$$ then $$ I$$  equals
    Solution
    As $$x $$ and $$\sqrt{1+x^{2}}$$ are involved,
    put $$ \sqrt{1-x^{2}} = t - x$$  or $$1 + x^{2}=t^{2} -2tx + x^{2}$$
    $$\displaystyle

    \Rightarrow  x= \frac{1}{2}\left(t-\frac{1}{t}\right) $$

    $$\displaystyle \Rightarrow dx = \frac{1}{2} \left

    (1+\frac{1}{t^{2}}\right )$$
    $$ \displaystyle \therefore I= \int_{0}^{\infty} \frac{1}{2} \left(1+\frac{1}{t^{2}}\right) \frac{dt}{t^{n}}$$
    $$=\displaystyle \frac{1}{2} \left.\left( \frac{t^{-n+1}}{1-n}-\frac{t^{-n-1}}{n+1}\right) \right ]_{1}^{\infty} $$
    $$=\displaystyle

    \frac{1}{2}\left. \left ( \frac{1}{1-n}\frac{1}

    {t^{n-1}}-\frac{1}{n+1}\frac{1}{t^{n+1}} \right) \right ]_{1}^{\infty}

    $$
    $$ =\displaystyle 0+\frac{1}{2} \left( \frac{1}{n-1}+\frac{1}{n+1}\right) =\frac{n}{n^{2}-1} $$
  • Question 7
    1 / -0
     If $$\displaystyle I = \int _{0}^{1} \frac{dx}{(1+ x)(2 + x)\sqrt{x(1-x)}}$$ then $$I$$ equals
    Solution
    Substitute $$ x =\sin^{2} \theta $$, so that

    $$\displaystyle I=\int_{0}^{\pi/2}\frac{2\sin \theta\cos \theta }{(1+\sin^{2} \theta )(2+\sin^{2} \theta )\sin \theta \cos \theta} d\theta $$

    $$\displaystyle = 2 \int_{0}^{\pi/2} \frac{\sec ^{4}\theta }{ (\sec^{2} \theta +\tan^{2} \theta )(2+\sec^{2} \theta +\tan^{2}\theta )} d\theta $$ 

    Substitute $$\tan \theta = t, $$ so that

    $$\displaystyle I= 2 \int_{0}^{\infty}\frac{(1+t^{2}) dt}{(1 + 2t^{2})(2+3t^{2})}$$

    $$\displaystyle =2\int_{0}^{\infty} \left[ \frac{1}{1+2t^{2}}-\frac{1}{2+3t^{2}} \right] dt $$

    $$\displaystyle =2 \left [ \frac{1}{\sqrt{2}}\tan^{-1} (\sqrt{2}t) -\frac{1}{\sqrt{2}\sqrt{3}}\tan^{-1}\left( \frac{\sqrt{3}t}{\sqrt{2}} \right)\right ]_ {0} ^{\infty} $$

    $$\displaystyle =\frac{\pi}{\sqrt{2}} \left( 1-\frac{1}{\sqrt{3}}\right) =\frac{\pi( \sqrt{3}-1)}{\sqrt{6}} $$
  • Question 8
    1 / -0
    If $$ h(x) = \int _{1}^{x} \sin^{4} t dt,$$ then $$ h(x + \pi)$$ equals
    Solution
     $$ h(x + \pi) = h(\pi) + I$$
    where $$ \displaystyle I=\int _{ \pi  }^{ x+\pi  }{ \sin ^{ 4 } tdt } $$
    Substitute $$ t = \theta + \pi$$ so that
    $$ I = \int _{0}^{x} (sin (\pi+ \theta )) ^{4} d\theta$$
    $$ =\int _{0}^{x} (-\sin \theta )^{4} d\theta =h(x) $$
    $$\therefore h(x + \pi) = h(\pi) + h(x)$$
  • Question 9
    1 / -0
     If $$\displaystyle \int _{0}^{1}\frac{\sin t}{1+t} dt = \alpha$$. then value of
    $$\displaystyle I=\int_{4\pi-2}^{4\pi}\frac{\sin(x/2)}{4\pi+2-x} dx $$
    Solution
    Substitute $$ \displaystyle \frac{x}{2}=\theta $$

    $$\displaystyle

    I=\int_{2\pi-1}^{2\pi} \frac{\sin \theta}{4\pi+2-2\theta} 2d\theta

    =\int_{2\pi-1}^{2\pi} \frac{\sin \theta}{2\pi+1-\theta} d\theta $$

    Now, substitute $$ 2\pi-\theta = t, $$ so that 

    $$\displaystyle I = \int_{1}^{0}\frac{ \sin(2\pi -t) }{1+t} (-1) dt $$ 

    $$=\displaystyle - \int_{0}^{1} \frac{\sin t}{1+t}dt=-\alpha $$
  • Question 10
    1 / -0
     If $$\displaystyle I = \int_{0}^{\pi/4}  \frac{ \sin 2\theta }{ \sin^{2} \theta +\cos ^{4} \theta } d \theta $$ 
    then $$I$$ is equal to?
    Solution
    $$ \sin^{2} \theta  + \cos^{4} \theta = (1/2) (1 -\cos 2\theta )+ (1/4) (1 + \cos 2\theta)^{2}$$

    $$ = (1/4) [2 -2 \cos 2\theta + 1+2 \cos 2\theta + cos^{2} 2\theta ] $$

    $$ = (1/4) (3 + \cos^{2} 2\theta)$$ 

    Substitute $$ \cos 2\theta  = t,$$ so that

    $$\displaystyle  I=\int_{1}^{0} \frac{-2 dt}{3+t^{2}}=\frac{2}{\sqrt{3}} \tan^{-1} \left.\left( \frac{t}{\sqrt{3}}\right)\right ]_{0}^{1}$$

    $$\displaystyle =\frac{2}{\sqrt{3}} \left(\frac{\pi}{6}\right) =\frac{\pi}{3\sqrt{3}}$$
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