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Integrals Test - 57

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Integrals Test - 57
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  • Question 1
    1 / -0
    $$\displaystyle\int^{\pi /3}_0\frac{\cos \theta}{5-4\sin \theta}d\theta$$ equal to.
    Solution
    $$I=\int _{ 0 }^{ { \pi  }/{ 3 } }{ \cfrac { \cos { \theta  }  }{ 5-4\sin { \theta  }  } d\theta  } $$

    Let $$5-4\sin { \theta =t } $$

    $$\Rightarrow \cos { \theta  } d\theta =-\cfrac { dt }{ 4 } $$
    $$I=\int _{ 5 }^{ 5-2\sqrt { 3 }  }{ \cfrac { -dt }{ 4t }  } $$

    $$I=\cfrac { 1 }{ 4 } \int _{ 5-2\sqrt { 3 }  }^{ 5 }{ \cfrac { dt }{ t }  } $$
    $$I=\cfrac { 1 }{ 4 } { \left[ \log { t }  \right]  }^{ 5 }_{ 5-2\sqrt { 3 }  }$$

    $$I=\cfrac { 1 }{ 4 } { \left[ \log { 5 } - \log { (5-2\sqrt { 3 } ) }  \right]  }$$

    $$I=\cfrac { 1 }{ 4 } { \left[ \log \left({ \cfrac { 5 }{ 5-2\sqrt { 3 }  }  }\right)  \right]  }$$

    $$\therefore I=\cfrac { 1 }{ 4 } \log { \left( \cfrac { 5 }{ 5-2\sqrt { 3 }  }  \right)  } $$

  • Question 2
    1 / -0
    Consider $$I=\displaystyle \int^{\pi}_{0}\displaystyle\frac{xdx}{1+\sin x}$$. What is I equal to?
    Solution
    $$I = \displaystyle \int_{0}^{\pi}{\dfrac{xdx}{1+\sin x}}$$

    by property,
    $$I = \displaystyle \int_{0}^{\pi}{\dfrac{(\pi-x)dx}{1+\sin x}}$$

    $$2I = \displaystyle \int_{0}^{\pi}{\dfrac{\pi dx}{1+\sin x}}$$

    $$2I =\displaystyle  \int_{0}^{\pi}{\dfrac{\pi }{1+\sin x}} \times \dfrac{1-\sin x}{1-\sin x}dx$$

    $$2I =\displaystyle  \int_{0}^{\pi}{\dfrac{\pi(1-\sin x) dx}{\cos^{2} x}}$$

    $$I = \dfrac{\pi}{2}\displaystyle \int _{0}^{\pi} [\sec^{2} x \tan x \sec x]dx$$


    $$I = \dfrac{\pi}{2} [\tan{x} - \sec{x}]_{0}^{\pi}$$

    $$I = \dfrac{\pi}{2} [0+2]$$

    $$I = \pi$$
  • Question 3
    1 / -0
    The value of $$\int _{ 1/3 }^{ 3 }{ \cfrac { \tan { \left( { x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } }  \right)  }  }{ \sin { \left( x+\cfrac { 1 }{ x }  \right)  }  }  } \cfrac { dx }{ x } $$ is 
  • Question 4
    1 / -0
    Let $$I_{1} =\displaystyle  \int_{0}^{1}\dfrac {e^{x}dx}{1 + x}$$ and $$I_{2} = \displaystyle \int_{0}^{1} \dfrac {x^{2}dx}{e^{x^{3}}(2 - x^{3})}$$, then $$\dfrac {I_{1}}{I_{2}}$$ is
    Solution
    Given : $${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) }  } \quad { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) }  } $$
    $${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) }  } \\ { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) }  } $$
    Let : $$1-{ x }^{ 3 }=t\quad x\rightarrow 0\quad t\rightarrow 1$$
    $$-3{ x }^{ 2 }=dt\quad x\rightarrow 1\quad t\rightarrow 0\\ { I }_{ 2 }=-\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 3 } \cfrac { { e }^{ t-1 } }{ (1+t) } dt } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt } \\ \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) }  }  }{ \cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt }  } =3e$$
    Hence the correct answer is $$3e$$
  • Question 5
    1 / -0
    $$\displaystyle\int^{\pi}_0\sqrt{|\cos x|-|\cos^3x|}dx$$ is?
    Solution
    Given : $$\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| -\left| \cos ^{ 3 }{ x }  \right| dx }  } $$
    $$I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| -\left| \cos ^{ 3 }{ x }  \right| dx }  } =\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right| \left| 1-\cos ^{ 2 }{ x }  \right| dx }  } \\ I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right|  } \left| \sin { x }  \right| dx } \\ I=\int _{ 0 }^{ \pi  }{ \sqrt { \left| \cos { x }  \right|  } \left( \sin { x }  \right) dx } \\ \sin { x } >0\quad (0<x<\pi )\\ \cos { x } =t\quad x\rightarrow 0\quad t\rightarrow 1\\ dt=-\sin { x } dx\quad x\rightarrow \pi \quad t\rightarrow -1\\ I=-\int _{ 1 }^{ -1 }{ \sqrt { \left| t \right|  } dt } \\ =\int _{ -1 }^{ 1 }{ \sqrt { \left| t \right|  } dt } \\ =\int _{ -1 }^{ 0 }{ \sqrt { -t } dt } +\int _{ 0 }^{ 1 }{ \sqrt { t } dt } \\ I=-\cfrac { 2 }{ 3 } \left| \left( -t \right) ^{ \cfrac { 3 }{ 2 }  } \right| _{ -1 }^{ 0 }+\cfrac { 2 }{ 3 } \left| t^{ \cfrac { 3 }{ 2 }  } \right| _{ 0 }^{ 1 }\\ I=-\cfrac { 2 }{ 3 } \left[ 0-1 \right] +\cfrac { 2 }{ 3 } (1-0)\\ =\cfrac { 4 }{ 3 } $$
    Hence the correct answer is $$\cfrac { 4 }{ 3 } $$
  • Question 6
    1 / -0
    The value of the definite integral $$\displaystyle\int^{\pi/2}_{0}\left(\cos ^{10}x\cdot \sin 12x\right)dx$$, is equal to.
    Solution
    Let $$ I =\displaystyle  \int_{0}^{\pi/2}(\cos^{10}x.\sin\, 12x) dx$$

    $$\cos^{10}x.\sin\,12x = \cos^{10}x.\sin(11x+x) = \cos^{10}x.(\sin\,11x.\cos\,x + \cos\, 11x.\sin\,x)$$
    $$= - \dfrac{1}{11}(-11\,\sin\,11x.\cos^{11}x - 11\,\cos\,11x.\cos^{10}x.\sin\,x)$$              
    $$= -\dfrac{1}{11} \times$$ $$\dfrac{d}{dx}(\cos\,11x.\cos^{11}x)$$ ... (1)
    Therefore, $$I = \left (-\dfrac {1}{11}\right)\displaystyle \int_{0}^{\pi/2} \dfrac{d}{dx}(\cos\,11x.\cos^{11}x) dx$$
    $$I =- \dfrac {1}{11}\times \cos\,11x.\cos^{11}x]_{0}^{\pi/2}$$
    $$I = -\dfrac {1}{11}\times (0-1)$$
    $$I = \dfrac{1}{11}$$
    Correct answer is option B.
  • Question 7
    1 / -0
    $$\int { { x }^{ 4 }{ e }^{ 2x } } dx=$$
    Solution
    $$\int x^4.e^{2x}dx$$

    Applying by parts rule of integration,

    $$\dfrac{x^4}{2} e^{2x}-\dfrac{4}{2}\int x^3.e^{2x}dx$$

    Applying by-parts again, multiple times till the time powers in $$x$$ in the integral vanishes,

    $$\dfrac{x^4}{2} e^{2x}-2[\dfrac{x^3}{2} e^{2x}-\dfrac{3}{2}\int x^2.e^{2x}dx]$$
    =>$$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3[\dfrac{x^2}{2}.e^{2x}-\int x.e^{2x} dx]$$
    =>$$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3[\dfrac{x}{2}.e^{2x}-\int\dfrac{e^{2x}}{2}dx]$$
    => $$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3\dfrac{x}{2}.e^{2x}+3\dfrac{e^{2x}}{4}+c$$
    $$\rightarrow   \cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$$
  • Question 8
    1 / -0
    The value of $$\int^3_{1/3}\dfrac{tan(x^2-\dfrac{1}{x^2})}{sin(x+\dfrac{1}{x}) }\dfrac{dx}{x}$$ is 
    Solution

  • Question 9
    1 / -0
    $$\displaystyle \int _{ 0 }^{ x }{ \cfrac { \sin { x }  }{ 1+\cos ^{ 2 }{ x }  }  } dx=\pi \cfrac { \cos { \alpha  }  }{ 1-\sin ^{ 2 }{ \alpha  }  } $$
    Solution

  • Question 10
    1 / -0
    $$\int _{ 0 }^{ \pi /2 }{ \sin ^{ 8 }{ x } \cos ^{ 2 }{ x } dx } $$ is equal to
    Solution

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