Self Studies
Self Studies
Self Studies
Self Studies

Integrals Test - 31

Result Self Studies

Integrals Test - 31
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    $$\displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\ dx$$ is equal to
    Solution
    Let $$I = \displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\, dx\, $$

    $$=\, \displaystyle \int_0^\pi \frac{1}{1\, +\,\displaystyle \frac{2 \tan \frac {x}{2}}{1\, +\, \tan^2 \frac {x}{2}}}\, dx$$

    $$=\, \displaystyle\int_0^\pi \frac{\sec^2 \dfrac {x}{2}}{\left(1\, +\,\tan \dfrac {x}{2}\right)^2}\, dx$$

    Put $$1+\tan$$ $$\dfrac{x}{2}\,= t\,\Rightarrow\,\dfrac{1}{2}\,\sec^2\,\dfrac{x}{2}\,=\,dt$$

    $$\therefore\,I\,=\,\displaystyle\int_1^\infty\dfrac{2dt}{1\,+\,t^2}\,=-\dfrac2t\bigg]_1^\infty = 2$$
  • Question 2
    1 / -0
    If $$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$$, then $$\int _{ 0 }^{ 2 }{ f(x)dx } $$ is
    Solution
     $$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$$ 

    from 0 to 1 we have $$2x^2+1$$ and from 1 to 2 we have $$4x^2-1$$

    $$=\int _{ 0 }^{ 1 }{ \left( 2{ x }^{ 2 }+1 \right) dx+\int _{ 1 }^{ 2 }{ \left( 4{ x }^{ 2 }-1 \right) dx }  } $$
    $$={ \left[ \frac { 2{ x }^{ 3 } }{ 3 } +x \right]  }_{ 0 }^{ 1 }+{ \left[ \frac { 4{ x }^{ 3 } }{ 3 } -x \right]  }_{ 1 }^{ 2 }$$
    $$=\frac { 2 }{ 3 } +1-0-0+\frac { 4{ (2) }^{ 3 } }{ 3 } -2-\left( \frac { 4 }{ 3 } -1 \right) $$
    $$=\frac { 2 }{ 3 } +1+\frac { 32 }{ 3 } -2-\frac { 4 }{ 3 } +1$$
    $$=\frac { 2+32-4 }{ 3 } $$
    $$=\dfrac{30}{3}$$
    =10.
  • Question 3
    1 / -0
    $$\displaystyle \int \frac{\sin 2x}{\sin^2x+2 \cos^2x}dx=$$
    Solution
    Let $$I = \displaystyle \int \frac{\sin 2x}{\sin^2x+2 \cos^2x}dx=\int \dfrac{\sin 2x}{(\sin^2x+\cos^2x)+\cos^2x}dx$$
    $$=\displaystyle \int \dfrac{\sin 2x}{1+\cos^2x}dx$$

    Substitute $$1+\cos^2x=u$$
    $$\Rightarrow 0+2\cos x(-\sin x)dx=du$$
    $$\Rightarrow \sin 2xdx=-du$$

    Therefore $$\displaystyle I=-\int \dfrac{du}{u}=-\ln u+c=-\ln(1+\cos^2x)+c$$
  • Question 4
    1 / -0
    If $$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$$, then $$b = $$
    Solution
    $$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$$
    $$\Rightarrow [\tan^{-1}x]_0^b=[\tan^{-1}x]_b^{\infty}$$
    $$\Rightarrow \tan^{-1}b-0=\dfrac{\pi}{2}-\tan^{-1}b$$
    $$\Rightarrow \tan^{-1}b=\dfrac{\pi}{4}$$
    $$\Rightarrow b=\tan\dfrac{\pi}{4}=1$$
  • Question 5
    1 / -0
    $$\displaystyle \int{\tan^2x}dx$$
    Solution
    $$\displaystyle \int{\tan^2x}dx=\int(\sec^2x-1)dx=\int \sec^2xdx-\int dx = \tan x-x+C$$
  • Question 6
    1 / -0
    The value of $$I=\displaystyle \int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \left( \tan ^{ n+1 }{ x }  \right) dx } +\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \left( \tan ^{ n+1 }{ \left( \frac { x }{ 2 }  \right)  }  \right) dx } $$ is
    Solution
    Given $$I=\displaystyle \int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \left( \tan ^{ n+1 }{ x }  \right) dx } +\cfrac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \left( \tan ^{ n+1 }{ \left( \cfrac { x }{ 2 }  \right)  }  \right) dx } $$ 

    In second integral, put $$t=\cfrac{x}{2} \Rightarrow$$ $$dx=2dt$$
    $$\Rightarrow$$ Also, when $$x=0$$ then $$t=0$$

    When $$x=\dfrac{\pi}2$$ then $$t=\dfrac{\pi}4$$
    Then $$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x }  \right)  } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ t }  \right)  } dt$$

    $$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x }  \right)  } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x }  \right)  } dx\quad (\because \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(y)dy } )$$

    $$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } +\tan ^{ n-1 }{ x }  \right)  } dx$$

    $$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x }  \right) .\left( \tan ^{ 2 }{ x+1 }  \right)  } dx$$

    $$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x }  \right) .\left( \sec ^{ 2 }{ x }  \right)  } dx$$

    Put $$t=\tan{x}$$
    $$\Rightarrow dt=\sec ^{ 2 }{ x } dx$$

    Also when $$x=0$$ then $$t=0$$
    when $$x=\pi /4$$, then $$t=1$$

    $$I=\displaystyle \int _{ 0 }^{ 1 }{ { t }^{ n-1 }dt } $$

    $$={ \left[ \cfrac { { t }^{ n } }{ n }  \right]  }_{ 0 }^{ 1 }=\cfrac { 1 }{ n } $$
  • Question 7
    1 / -0
    $$\displaystyle \int (1 + x - x^{-1})e^{x + x^{-1}}dx$$ is equal to
    Solution
    $$\displaystyle \int (1 + x - x^{-1})e^{x + x^{-1}}dx$$

    $$\displaystyle = \int [x.e^{x + x^{-1}} \left (1 - \dfrac {1}{x^{2}}\right ) + e^{x + x^{-1}}]dx$$       $$\left [\displaystyle \because \int xf'(x) + f(x) dx = xf(x) + C \right]$$

    $$\displaystyle \therefore \int (1 + x - x^{-1})e^{x + x^{-1}} dx = xe^{x + x^{-1}} + C$$
  • Question 8
    1 / -0
    What is $$\displaystyle \int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ 1+{ x }^{ 2 } }  } dx$$ equal to?
    Solution
    $$I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { \tan ^{ -1 }{ x }  }{ 1+{ x }^{ 2 } }  } dx$$ 
    Put $$\tan ^{ -1 }{ x } =p\Rightarrow \dfrac { dx }{ 1+{ x }^{ 2 } } =dp$$
    When $$x=0, p=0$$ and $$x=1,p=\dfrac{\pi}{4}$$
    $$ \Rightarrow I=\displaystyle \int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ pdp }=\dfrac{p^{2}}{2} =\dfrac { { \pi  }^{ 2 } }{ 32} $$
    Hence, B is correct.
  • Question 9
    1 / -0
    $$\displaystyle\int^{\pi /2}_{-\pi /2}\cos x$$ ln $$\left(\displaystyle\frac{1+x}{1-x}\right)dx$$ is equal to.
    Solution
    Let the given integral be $$I$$

    We have $$I=\displaystyle \int _{ -\tfrac { \pi  }{ 2 }  }^{ \tfrac { \pi  }{ 2 }  }{\cos x\ln { \left(\dfrac { 1+x }{ 1-x }\right)dx }} $$
    In the above expression , $$x$$ can be replaced with $$\dfrac{\pi}{2}-\dfrac{\pi}{2}-x=-x$$

    $$\Rightarrow I=\displaystyle \int _{ -\tfrac { \pi  }{ 2 }  }^{ \tfrac { \pi  }{ 2 }  }{ \cos x\ln { \left(\dfrac { 1+x }{ 1-x } \right)dx } =\int _{ -\tfrac { \pi  }{ 2 }  }^{ \tfrac { \pi  }{ 2 }  }{\cos x\ln { \left(\dfrac { 1-x }{ 1+x }\right)dx } \quad  } \quad \quad  } $$

    By adding both, we get $$2I=\displaystyle \int _{ -\tfrac { \pi  }{ 2 }  }^{ \tfrac { \pi  }{ 2 }  } \cos x\left(\ln \left(\dfrac { 1+x }{ 1-x } \right)+\ln { \left(\dfrac { 1-x }{ 1+x } \right) } \right)dx$$ 

    $$=\displaystyle \int _{ -\tfrac { \pi  }{ 2 }  }^{ \tfrac { \pi  }{ 2 }  }  \cos x\left(\ln \left(\dfrac { 1+x }{ 1-x } \times \dfrac { 1-x }{ 1+x } \right)\right)dx =\int _{-\tfrac {\pi}{2}}^{\tfrac{\pi}{2}} \cos x(\ln (1))dx =0$$

    $$\Rightarrow I=0$$
    Therefore the correct option is $$A$$
  • Question 10
    1 / -0
    What is $$\displaystyle\int _{ 1 }^{ 2 }{ \ln { x } dx } $$ equal to?
    Solution
    Let, $$y=\ln x$$ then, $$ x={ e^{ y } }$$
    Now at, $$ x=1,y=0$$
                 $$ x=2,y=\ln 2$$
    $$\Rightarrow dx={ e }^{ y }.dy$$

    Putting the values, we get

    $$\displaystyle \int_{1}^{2} \ln x dx=\int _{ 0 }^{ \ln2 }{ { e }^{ y }.y.dy }$$

     $$ =\displaystyle { \left[ y.{ e }^{ y } \right]  }_{ 0 }^{\ln2 }-\int _{ 0 }^{\ln2 }{ { e }^{ y }dy }$$ 

    $$ =2\ln 2-1$$ 

    $$=2\ln 2-\ln(e)$$

    $$ =\ln\left(\dfrac { 4 }{ e }\right)$$ 
    Hence, C is correct.
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Selfstudy
Selfstudy
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now