Integrals Test - 33

Let $$I=\displaystyle \int_{\log 2}^{x}\frac{1}{\sqrt{e^x-1}}dx$$
Put $$e^x-1=t^2$$ $$\Rightarrow e^xdx=2t\, dt \Rightarrow (t^{2}+1)dx=2t\ dt \Rightarrow dx=\dfrac{2t}{1+t^{2}} dt$$

Let $$I=\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right)  } dx } $$
$$=\cfrac { 1 }{ 7 } \int { \sin { \left( \cfrac { x }{ 7 } +10 \right)  }  } =\cfrac { 1 }{ 7 } \cfrac { -\cos { \left( \cfrac { x }{ 7 } +10 \right)  }  }{ \cfrac { 1 }{ 7 }  } $$
$$=-\cos { \left( \cfrac { x }{ 7 } +10 \right)  } +C$$

$$\int { \sqrt { 1+\sin { x }  }  } .f\left( x \right) dx=\dfrac { 2 }{ 3 } { (1+\sin { x } ) }^{ { 3 }/{ 2 } }+c$$

Let $$I=\displaystyle\int { { x }^{ x }\left( \log { ex }  \right) dx } $$
$$=\displaystyle\int { { x }^{ x }\left( 1+\log { x }  \right) dx } $$
Let $$t={ x }^{ x }={ e }^{ x\log { x }  }$$
$$\Rightarrow \dfrac { dt }{ dx } ={ x }^{ x }\left\{ x\cdot \dfrac { 1 }{ x } +\log { x }  \right\} $$
$$\Rightarrow dt={ x }^{ x }\left( 1+\log { x }  \right) dx$$
Therefore, $$ I=\displaystyle\int { t+c } ={ x }^{ x }+c$$

$$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ a+b\cos { x }  }  } =\int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ a+b\left(\dfrac { 1-\tan ^{ 2 }{ \dfrac { x }{ 2 }  }  }{ 1+\tan ^{ 2 }{ \dfrac { x }{ 2 }  }  } \right) }  } =\int _{ 0 }^{ \pi /2 }{ \dfrac { (1+\tan ^{ 2 }{ \dfrac { x }{ 2 }  } )dx }{ (a+b)+(a-b)\tan ^{ 2 }{ \dfrac { x }{ 2 }  }  }  } =\int _{ 0 }^{ \pi /2 }{ \dfrac { \sec ^{ 2 }{ \dfrac { x }{ 2 }  } dx }{ (a+b)+(a-b)\tan ^{ 2 }{ \dfrac { x }{ 2 }  }  }  } $$

$$I=\displaystyle\int _{ 0 }^{ { \sqrt { \pi  }  }/{ 2 } }{ 2{ x }^{ 3 }\sin { \left( { x }^{ 2 } \right)  } dx } $$
   $$=\displaystyle\int _{ 0 }^{ { \sqrt { \pi  }  }/{ 2 } }{ 2{ x }^{ 2 }\cdot x\sin { \left( { x }^{ 2 } \right)  } dx } $$
Put $${ x }^{ 2 }=t$$
$$\Rightarrow 2xdx=dt$$
Also, when $$x=0$$, then $$t=0$$
and when $$x=\dfrac { \sqrt { \pi  }  }{ 2 } $$, then $$t=\dfrac { \pi  }{ 4 }$$
$$\Rightarrow I=\displaystyle\int _{ 0 }^{ { \pi  }/{ 4 } }{ \underset { I }{ t } \underset { II }{ \sin { t }  } dt } $$
     $$={ \left[ t\left( -\cos { t }  \right)  \right]  }_{ 0 }^{ { \pi  }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi  }/{ 4 } }{ -\cos { t } \left( 1 \right) dt } $$
     $$={ \left[ -t\cos { t }  \right]  }_{ 0 }^{ { \pi  }/{ 4 } }+\displaystyle\int _{ 0 }^{ { \pi  }/{ 4 } }{ \cos { t } dt } $$
     $$={ \left[ -t\cos { t } +\sin { t }  \right]  }_{ 0 }^{ { \pi  }/{ 4 } }$$
     $$=\left[ -\dfrac { \pi  }{ 4 } \cdot \dfrac { 1 }{ \sqrt { 2 }  } +\dfrac { 1 }{ \sqrt { 2 }  }  \right] $$
     $$=\dfrac { 1 }{ \sqrt { 2 }  } \left( 1-\dfrac { \pi  }{ 4 }  \right) $$

$$1+{ x }^{ 2 }=t$$
$$2xdx=dt$$
$$xdx=\cfrac { dt }{ 2 } $$
$$x=0,t=1$$
$$x=\infty ,t=\infty $$
$$x=\sqrt { t-1 } $$
$$\ln { x } =\cfrac { 1 }{ 2 } \ln { \left( t-1 \right)  } $$
$$\Rightarrow \int _{ 1 }^{ \infty  }{ \cfrac { 1 }{ 2 } \cfrac { \ln { \left( t-1 \right)  }  }{ { t }^{ 2 } }  } \cfrac { dt }{ 2 } $$
$$=\cfrac { 1 }{ 4 } \int _{ 1 }^{ \infty  }{ \cfrac { 1 }{ { t }^{ 2 } } \ln { \left( t-1 \right)  }  } dt$$
Integrating by parts
$$=\cfrac { 1 }{ 4 } { \left[ -\cfrac { 1 }{ t } \ln { \left( t-1 \right)  }  \right]  }_{ 1 }^{ \infty  }-\cfrac { 1 }{ 4 } \int _{ 1 }^{ \infty  }{ -\cfrac { 1 }{ t\left( t-1 \right)  }  } dt$$
$$=\cfrac { 1 }{ 4 } { \left[ -\cfrac { 1 }{ \left( 1+{ x }^{ 2 } \right)  } \ln { \left( { x }^{ 2 } \right)  }  \right]  }_{ 0 }^{ \infty  }+\cfrac { 1 }{ 4 } \left[ \int _{ 1 }^{ \infty  }{ \cfrac { 1 }{ t-1 }  } dt-\int _{ 1 }^{ \infty  }{ \cfrac { 1 }{ t-1 }  } dt \right] $$
$$=\cfrac { 1 }{ 4 } \times 0+\cfrac { 1 }{ 4 } \times 0=0$$

$$\int_{1}^{2} e^{x^{2}} dx = a$$ (given)
Let $$I = \int_{e}^{e^{4}}\sqrt {\log x} dx$$ Putting $$\log x = t^{2} \ dx = 2te^{t^{2}} dt$$
$$= \int_{1}^{2} t(2t)e^{t^{2}} dt = \int_{1}^{2} t(2t e^{t^{2}}) dt$$
$$= [te^{t^{2}}]_{1}^{2} - \int_{1}^{2} e^{t{2}} dt = 2e^{4} - e - a$$
$$\left \{As \int_{1}^{2} e^{t^{2}} dt = \int_{1}6{2} e^{x^{2}} dx = a\right )$$
Hence (a) is correct choice.

Given that
$$f\left( x \right) =4{ x }^{ 2 }-3x+1,$$ $$g\left( x \right) =\dfrac { f\left( -x \right) -f\left( x \right)  }{ { x }^{ 2 }+3 } $$
Therefore, $$ g\left( x \right) =\dfrac { \left( 4{ x }^{ 2 }+3x+1 \right) -\left( 4{ x }^{ 2 }-3x+1 \right)  }{ { x }^{ 2 }+3 } $$
$$=\dfrac { 6x }{ { x }^{ 2 }+3 } $$
Now, $$g\left( -x \right) =-\dfrac { 6x }{ { x }^{ 2 }+3 } =-g\left( x \right)$$
Which is an odd function
Thus $$ \displaystyle\int _{ -2 }^{ 2 }{ g\left( x \right) dx } =0$$