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Integrals Test - 27

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Integrals Test - 27
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  • Question 1
    1 / -0
    $$\displaystyle \int_{\pi /4}^{3\pi /4}\frac{dx}{1+\cos x}$$ is equal to
    Solution
    Let $$\displaystyle I=\int { \dfrac { 1 }{ 1+\cos { x }  }  } dx$$

    Substitute $$\displaystyle t=\tan { \dfrac { x }{ 2 }  } \Rightarrow dt=\dfrac { 1 }{ 2 } dxsec^{ 2 }x$$

    $$\displaystyle I=\int { \dfrac { 2 }{ \left( { u }^{ 2 }+1 \right) \left( \dfrac { 1-{ u }^{ 2 } }{ { u }^{ 2 }+1 } +1 \right)  }  } du=\int { du } =u=\tan { \dfrac { x }{ 2 }  } $$

    $$\displaystyle \therefore \int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { dx }{ 1+\cos { x }  }  } =\left[ \tan { \dfrac { x }{ 2 }  }  \right] _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }=2$$
  • Question 2
    1 / -0
    If $$I_1 = \displaystyle \int_x^1 \frac{1}{1 + t^2} dt$$ and $$I_2 \displaystyle = \int_1^{1 / x}\frac{1}{1+ t^2}dt$$ for $$x > 0$$, then
    Solution
    $${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } dt } $$
    Substitute $$t=\dfrac { 1 }{ u } \Rightarrow dt=-\dfrac { 1 }{ u^{ 2 } } du$$
    We get
    $$I_{ 1 }=\int _{ 1/x }^{ 1 } \dfrac { 1 }{ 1+\dfrac { 1 }{ u^{ 2 } }  } \left( -\dfrac { 1 }{ u^{ 2 } }  \right) du=-\int _{ 1/x }^{ 1 } \dfrac { du }{ u^{ 2 }+1 } =\int _{ 1 }^{ 1/x } \dfrac { 1 }{ 1+u^{ 2 } } du=I_{ 2 }$$
  • Question 3
    1 / -0
    If $$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$$ for $$x> 0,$$ 
    and $$\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$$ 
    then $$\displaystyle I$$ is?
    Solution
    Substitute $$\displaystyle z=\dfrac { 1 }{ t } $$

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

    $$\displaystyle =\int _{ x }^{ \dfrac { 1 }{ x }  }{ f\left( t \right)  } dt=-\int _{ \dfrac { 1 }{ x }  }^{ x }{ f\left( t \right)  } dt=-I\\ \therefore 2I=0$$
  • Question 4
    1 / -0
    $$ \displaystyle \int_{\sin \theta }^{\cos \theta }f(x \tan \theta )dx\left ( where \theta \neq \frac{n\pi}{2} ,n\epsilon I\right )$$ is equal to
    Solution
    Let $$I=\int _{ sin\theta  }^{ cos\theta  }{ f\left( xtan\theta  \right) dx } $$
    Substitute $$xtan\theta =zsin\theta \Rightarrow dx=cos\theta dz$$
    Therefore $$x=sin\theta \Rightarrow z=tan\theta $$ and $$x=cos\theta \Rightarrow z=1$$
    Gives
    $$I=\int _{ tan\theta  }^{ 1 }{ f\left( zsin\theta  \right) dz } =-\int _{ 1 }^{ tan\theta  }{ f\left( zsin\theta  \right) dz } =-\int _{ 1 }^{ tan\theta  }{ f\left( xsin\theta  \right) dx } $$
  • Question 5
    1 / -0
    Suppose that F(x) is an anti-derivative of $$\displaystyle f(x)=\frac{\sin x}{x}$$, where $$x>0$$.
    Then $$\displaystyle \int_{1}^{3}\dfrac{\sin2x}{x} \:dx$$ can be expressed as?
    Solution
    Let $$I=\displaystyle \int _{ 1 }^{ 3 }{ \cfrac { sin2x }{ x } dx } $$
    Substitute $$2x=t\Rightarrow 2dx=dt$$
    Gives $$I=\displaystyle \int _{ 2 }^{ 6 }{ \cfrac { sint }{ t } dt } $$
    As antiderivative of $$f\left( x \right) =\cfrac { sinx }{ x } $$ is $$F\left( x \right) $$
    $$I={ \left[ F\left( x \right)  \right]  }_{ 2 }^{ 6 }=F\left( 6 \right) -F\left( 2 \right) $$
  • Question 6
    1 / -0
    If $$\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$$ then $$\displaystyle {f}'\left ( \frac{\pi }{3} \right )$$ is
    Solution
    $$\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x }  }{ 1+{ t }^{ 2 } }  } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } }  } dt$$
    $$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t }  \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi  }{ 4 } +\frac { \pi  }{ 4 }  \right] =\sin { x } \left[ \frac { \pi  }{ 2 }  \right] $$
    $$\displaystyle \therefore f'\left( x \right) =\frac { \pi  }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi  }{ 3 }  \right) =\frac { \pi  }{ 2 } \frac { 1 }{ 2 } =\frac { \pi  }{ 4 } $$
  • Question 7
    1 / -0
    The solution for x of the equation $$\displaystyle \int_{\sqrt{2}}^{x}\frac{dt}{t\sqrt{t^{2}-1}}=\frac{\pi }{2}$$ is
    Solution
    As we know,

     $$\int{\dfrac{dt}{t\sqrt{t^2-1}}}= \left[\sec^{-1}t\right]+C$$


    $$\therefore \displaystyle \int_{\sqrt 2}^{x}{\dfrac{dt}{t\sqrt{t^2-1}}}=\dfrac{\pi}{2}$$

    $$\Rightarrow \left[\sec^{-1}t\right]_{\sqrt 2}^{x}=\dfrac{\pi}{2}$$

    $$\Rightarrow [\sec^{-1}x-\sec^{-1}\sqrt{2}]=\dfrac{\pi}{2}$$.

    $$\Rightarrow \sec^{-1}x=\sec^{-1}\sqrt 2+\dfrac{\pi}{2}$$

                        $$=\dfrac{\pi}{4}+\dfrac{\pi}{2}$$

                        $$=\dfrac{3\pi}{4}$$

    $$\Rightarrow x=\sec\left(\dfrac{3\pi}{4}\right)=\dfrac{1}{\cos \left(\dfrac{3\pi}{4}\right)}=\dfrac{1}{\cos(\pi-\dfrac{\pi}4)}=\dfrac{1}{-\cos \dfrac {\pi}4}$$

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

                                                                                          $$=-\sqrt 2$$


    $$\therefore \boxed{x=-\sqrt 2}\rightarrow (B)$$

  • Question 8
    1 / -0
    $$\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx$$
    Solution
    $$ \displaystyle \int _{ 0 }^{ 1 } \frac { 2^{ x+1 }-3^{ x-1 } }{ 6^{ x } } dx=\int _{ 0 }^{ 1 } \left( 2.3^{ -x }-\frac { 1 }{ 3 } 2^{ -x } \right) dx$$

    $$ \displaystyle =\left[ -2\frac { 3^{ -x } }{ \log  3 } +\frac { 1 }{ 3 } \frac { 2^{ -x } }{ \log  2 }  \right] ^{ 1 }$$

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

    $$\displaystyle =\frac { 4 }{ 3 } \log _{ 3 } e-\frac { 1 }{ 6 } \log _{ 2 } e$$
    Hence, option 'A' is correct.
  • Question 9
    1 / -0
    $$\displaystyle \int_{1}^{2}\left ( x+\frac{1}{x} \right )^{3/2}\frac{x^{2}-1}{x^{2}}dx$$
    Solution
    Let $$\displaystyle I=\int _{ 1 }^{ 2 } \left( x+\frac { 1 }{ x }  \right) ^{ 3/2 }\frac { x^{ 2 }-1 }{ x^{ 2 } } dx$$

    Put $$\displaystyle x+\frac { 1 }{ x } =t\Rightarrow \left( 1-\frac { 1 }{ { x }^{ 2 } }  \right) dx\Rightarrow dt$$
    Therefore
    $$\displaystyle I=\int _{ 2 }^{ 5/2 } t^{ 3/2 }dt=\frac { 2 }{ 5 } t^{ 5/2 }=\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 }  \right) ^{ 5/2 }-2^{ 5/2 } \right] $$

    $$\displaystyle =\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 }  \right) ^{ 2 }\sqrt { \left( \frac { 5 }{ 2 }  \right)  } -2^{ 2 }\sqrt { 2 }  \right] =\frac { 5 }{ 2 } \sqrt { \left( \frac { 5 }{ 2 }  \right)  } -\frac { 8 }{ 5 } \sqrt { 2 } $$
    Hence, option 'B' is correct.
  • Question 10
    1 / -0
    The value of$$\displaystyle \int_{1}^{2}\frac{\cos \left ( \log x \right )}{x}dx$$  is equal to
    Solution
    $$\displaystyle \int _{ 1 }^{ 2 } \dfrac { \cos  \left( \log  x \right)  }{ x } dx$$

    Substitute $$\log  x=t$$
    $$\displaystyle \frac { 1 }{ x } dx=dt$$

    $$\displaystyle I=\displaystyle \int _{ \log { 1 }  }^{ \log { 2 }  } \cos { t } dt$$

    $$\displaystyle I={ [\sin { t } ] }_{ \log { 1 }  }^{ \log { 2 }  }$$

    $$\displaystyle I=\sin { (\log { 2) }  } $$
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