Integrals Test - 34

$$I = \int \dfrac{(x + 1 - 1)e^x}{(1 + x)^2} dx$$

Let $$I=\int_{-4}^{4}f(x^2)dx$$

$$\therefore I=\int_{-4}^{0}f(x^2)dx+\int_{0}^{4}f(x^2)dx$$

$$\therefore I=\int_{4}^{0}f((-x)^2)dx+\int_{0}^{4}f(x^2)dx$$

$$\therefore I=2\int_{0}^{4}f(x^2)dx$$

$$\therefore I=2\int_{0}^{4}(4x^3-g(4-x))dx$$ (from given equation)

$$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{4}g(4-x)dx$$

$$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{2}^{4}g(4-x)dx$$

$$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{2}^{4}g(4-x)dx$$ ... (1)

Let $$I_1=\int_{2}^{4}g(4-x)dx$$

Now, consider $$(4-x)=y$$ which is substituted in $$I_1$$ 

$$\therefore I_1=\int_{0}^{2}g(y)dy$$ (inverting limits and $$dy$$ both provide negative signs)

Substituting $$I_1$$ in (1), we have

$$ I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{0}^{2}g(y)dy$$

Also, $$g(4-x)=-g(x)$$ from given equation and $$g(y)=g(x)$$ from variable substitution

$$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}-g(x)dx-2\int_{0}^{2}g(x)dx$$

$$\therefore I=2\int_{0}^{4}4x^3dx$$

$$\therefore I=512$$

This is the required answer.

$$I=\int { \cfrac { 1 }{ x\sqrt { 1-{ x }^{ 3 } }  } dx } =\int { \cfrac { { x }^{ 2 } }{ { x }^{ 3 }\sqrt { 1-{ x }^{ 3 } }  } dx } $$

Let $$\sqrt { 1-{ x }^{ 3 } } =t$$

$$I=\int { \cfrac { 1 }{ x\sqrt { 1-{ x }^{ 3 } }  } dx } =\int { \cfrac { { x }^{ 2 } }{ { x }^{ 3 }\sqrt { 1-{ x }^{ 3 } }  } dx } \\ \Rightarrow \cfrac { 1 }{ 2\sqrt { 1-{ x }^{ 3 } }  } x(-3{ x }^{ 2 })dx=dt\\ \Rightarrow I=\cfrac { 2 }{ 3 } \int { \cfrac { 1 }{ { t }^{ 2 }-1 } dt } \\ \Rightarrow \cfrac { 3 }{ 2 } I=\int { \cfrac { 1 }{ (t+1)(t-1) } dt } =\cfrac { 1 }{ 2 } \int { \cfrac { (t+1)-(t-1) }{ (t+1)(t-1) }  } dt\\ \Rightarrow 3I=\int { \cfrac { 1 }{ (t-1) } dt } -\int { \cfrac { 1 }{ (t+1) } dt } \\ =\ln { \left| t-1 \right|  } -\ln { \left| t+1 \right|  } +b$$

where $$b$$ is a constant

$$\Rightarrow 3I= \ln{ \left| \cfrac{t-1}{t+1} \right| }+b$$

Converting $$t=\sqrt{1-x^3}$$, we get

$$I=\cfrac{1}{3} \ln{ \left| \cfrac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} +1} \right| }$$

Hence, the correct answer is $$a=\cfrac{1}{3}$$

$$I=\displaystyle \int_{1}^{\infty}\dfrac{e^x}{e^{2x+1}+e^{3}}dx=\dfrac{1}{e}\int_{1}^{\infty}\dfrac{e^x}{e^{2x}+e^{2}}dx$$

Substitute $$e^x=t\Rightarrow e^xdx=dt$$ and limits will be 

Lower limit $$x=1\Rightarrow t=e$$ 

Upper limit $$x=\infty\Rightarrow t=\infty$$

$$I=\displaystyle \dfrac{1}{e}\int_{e}^{\infty}\dfrac{1}{t^{2}+e^{2}}dt$$ 

$$=\dfrac{1}{e}\left| \dfrac{1}{e}\tan^{-1}\left | \dfrac{t}{e}\right |\right |_{e}^{\infty}=\dfrac{1}{e^2}\left | \dfrac{\pi}{2}-\dfrac{\pi}{4}\right |=\dfrac{\pi}{4e^2}$$

Hence, $$(A)$$

Let

$$\begin{matrix} I=\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sqrt { \sin  \phi  } { { \cos   }^{ 5 } }\phi d\phi  }  \\ =\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sqrt { \sin  \phi  } { { \cos   }^{ 4 } }\phi \cos  \phi d\phi  }  \\ =\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sqrt { \sin  \phi  } { { \left( { 1-{ { \sin   }^{ 2 } }\phi  } \right)  }^{ 2 } }\cos  \phi d\phi  }  \\ put\, \, \, t=\sin  \phi  \\ \dfrac { { dt } }{ { d\phi  } } =\cos  \phi  \\ dt=\cos  \phi d\phi  \\ When\, \, \phi =0,\, \, t=0\, \, when\, \, \phi =\dfrac { \pi  }{ 2 } ,\, t=1 \\ \therefore I=\int _{ 0 }^{ 1 }{ \sqrt { t } { { \left( { 1-{ t^{ 2 } } } \right)  }^{ 2 } }dt }  \\ =\int _{ 0 }^{ 1 }{ \sqrt { t } \left( { 1-2{ t^{ 2 } }+{ t^{ 4 } } } \right) dt }  \\ =\int _{ 0 }^{ 1 }{ { t^{ \dfrac { 1 }{ 2 }  } }+{ t^{ \dfrac { 9 }{ 2 }  } }-2{ t^{ \dfrac { 5 }{ 2 }  } }dt }  \\ =\left[ { \dfrac { 2 }{ 3 } { t^{ \dfrac { 3 }{ 2 }  } }+\dfrac { 2 }{ { 11 } } { t^{ \dfrac { { 11 } }{ 2 }  } }-\dfrac { 4 }{ 7 } { t^{ \dfrac { 7 }{ 2 }  } } } \right] _{ 0 }^{ 1 } \\ =\dfrac { 2 }{ 3 } +\dfrac { 2 }{ { 11 } } -\dfrac { 4 }{ 7 }  \\ =\dfrac { { 154+42-132 } }{ { 3\times 11\times 7 } } =\dfrac { { 64 } }{ { 231 } }  \\  \end{matrix}$$

Let 

      $$\begin{matrix} I=\int _{ 0 }^{ \infty  }{ \dfrac { { x{ { \tan   }^{ -1x } } } }{ { \left( { 1+{ x^{ 2 } } } \right) { x^{ 2 } } } } dx }  \\ \, \, \, \, \, \, =\int _{ 0 }^{ \infty  }{ \dfrac { { { { \tan   }^{ -1x } } } }{ { \left( { 1+{ x^{ 2 } } } \right) x } } dx }  \\  \end{matrix}$$

Now,

    $$\begin{matrix} \dfrac { { dI } }{ { dx } } =\int _{ 0 }^{ \infty  }{ \dfrac { 1 }{ { x\left( { 1+{ x^{ 2 } } } \right)  } } \left[ { \dfrac { 1 }{ { 1+{ x^{ 2 } } } } .xdx } \right]  }  \\ \, \, \, \, \, \, \, \, =\int _{ 0 }^{ \infty  }{ \dfrac { { dx } }{ { \left( { 1+{ x^{ 2 } } } \right) \left( { 1+{ x^{ 2 } } } \right)  } }  } \\, \, \, \quad \quad \quad \quad \quad =\int _{ 0 }^{ \infty  }{ \left[ { \dfrac { 1 }{ { 1+{ x^{ 2 } } } } -\dfrac { 1 }{ { 1+{ x^{ 2 } } } }  } \right]  } dx \\ \, \, \, \, \, \, \, \, \, =\dfrac { \pi  }{ 2 } \log  2 \\  \end{matrix}$$

Hence option (a) is correct.

Consider $$\displaystyle \int_{0}^{\pi/4}\cos^3 (2x)\:dx$$

     $$\begin{matrix} I=\int _{ 0 }^{ 1 }{ \frac { { 8\log  \left( { 1+x } \right)  } }{ { 1+{ x^{ 2 } } } }  } dx \\ \, \, \, \, =8\int _{ 0 }^{ 1 }{ \frac { { \log  \left( { 1+x } \right)  } }{ { 1+{ x^{ 2 } } } }  } dx \\ put\, \, \, x=\tan  \theta  \\ \, \, \, \, \, \Rightarrow dx={ \sin ^{ 2 }  }\theta d\theta  \\ x,\, \frac { \pi  }{ 4 } ,\, \, \, \, x=0 \\ \quad \quad \quad \quad I=8\int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \frac { { \log  \left( { 1+x } \right)  } }{ { { { \sin   }^{ 2 } }\theta  } }  } { \sin ^{ 2 }  }\theta d\theta  \\ \, \, \, \, \, \quad \quad \quad \quad =8\int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \log  \left( { 1+\tan  x } \right)  } dx \\ \quad \quad =8\times \frac { \pi  }{ 8 } \log  2 \\ \, \, \, =\pi \log  2 \\  \end{matrix}$$

Hence option (A) is true.