Integrals Test - 30

$$\displaystyle F\left ( x \right )=\int \frac{\sin x}{x}dx$$ Now $$\displaystyle I = \int_{1}^{3}\frac{\sin 2x}{x}dx\left [ put2x=t \right ]=\int_{2}^{6}\frac{2}{2}\frac{\sin }{t}dt=\left [ F\left ( x \right ) \right ]_{2}^{6}=F(6)-F(2)$$

 $$\displaystyle I =\int_{-1}^{1} (1 -2\alpha x + \alpha^{2} )^{-1/2}  dx$$
$$\displaystyle =\left. \frac{2(1-2\alpha x+\alpha^{2})^{1/2}}{-2\alpha}\right ]_{-1}^{1}$$
$$\displaystyle =-\frac{1}{\alpha}[([(1 -2\alpha + \alpha^{2})^{1/2} -(1 + 2 \alpha + \alpha ^{2})^{1/2} ]$$

$$\displaystyle I=\int _{ \frac { 1 }{ 2 }  }^{ 2 }{ \frac { 1 }{ x }  } \sin { \left( x-\frac { 1 }{ x }  \right) dx } $$

Put $$\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=\frac { 1 }{ { t }^{ 2 } } dt$$

$$\displaystyle I=\int _{ 2 }^{ \frac { 1 }{ 2 }  }{ t\sin { \left( \frac { 1 }{ t } -t \right) \left( \frac { -1 }{ { t }^{ 2 } }  \right) dt }  } =\int _{ 2 }^{ \frac { 1 }{ 2 }  }{ \frac { 1 }{ t } \sin { \left( t-\frac { 1 }{ t }  \right) dt }  } $$

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

$$\Rightarrow 2I=0\quad \Rightarrow I=0$$

$$I=\int _{ 1 }^{ \infty  }{ \cfrac { 1 }{ { e }^{ x+1 }+{ e }^{ 3-x } } dx } =\int _{ 1 }^{ \infty  }{ \cfrac { { e }^{ x-1 } }{ \left( { e }^{ x }-ie \right) \left( { e }^{ x }+ie \right)  } dx } $$
Substituting $${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$$
$$I=\cfrac { 1 }{ e } \int _{ e }^{ \infty  }{ \cfrac { 1 }{ \left( t-ie \right) \left( t+ie \right)  } dt } =\cfrac { 1 }{ e } \int _{ e }^{ \infty  }{ \cfrac { 1 }{ \left( e-it \right) \left( e+it \right)  } dt } \\ =\cfrac { 1 }{ e } \int _{ e }^{ \infty  }{ \left( \cfrac { i }{ 2\left( et+i{ e }^{ 2 } \right)  } -\cfrac { i }{ 2\left( et-i{ e }^{ 2 } \right)  }  \right)  } dt\\ ={ \left[ \cfrac { i\log { \left( et+i{ e }^{ 2 } \right)  }  }{ 2{ e }^{ 2 } } -\cfrac { i\log { \left( et-i{ e }^{ 2 } \right)  }  }{ 2{ e }^{ 2 } }  \right]  }_{ e }^{ \infty  }\\ ={ \left[ \cfrac { \tan ^{ -1 }{ \left( { e }^{ x-1 } \right)  }  }{ { e }^{ 2 } }  \right]  }_{ e }^{ \infty  }=\cfrac { \pi  }{ 4{ e }^{ 2 } } $$

Let $$\displaystyle I=\int { \frac { \sin { x }  }{ \sin { x } +\cos { x }  } dx } $$
Multiply numerator and denominator by $$I=\int { \dfrac { \sin { x }  }{ \sin { x } +\cos { x }  } dx } \\ \csc ^{ 3 }{ x } $$ 
$$\displaystyle I=\int { \frac { \csc ^{ 2 }{ x }  }{ \csc ^{ 2 }{ x } +\cot { x } \csc ^{ 2 }{ x }  } dx } =\int { \frac { \csc ^{ 2 }{ x }  }{ \cot ^{ 3 }{ x } +\cot ^{ 2 }{ x } +\cot { x } +1 } dx } $$

Put $$t=\cot { x } \Rightarrow dt=-\csc ^{ 2 }{ x } dx$$
$$\displaystyle I=-\int { \frac { 1 }{ { t }^{ 3 }+{ t }^{ 2 }+t+1 } dt } =-\int { \frac { 1-t }{ 2\left( { t }^{ 2 }+1 \right)  } dt } -\int { \frac { 1 }{ 2\left( t+1 \right)  } dt } $$

$$\displaystyle =\frac { 1 }{ 2 } \int { \frac { t }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ t+1 } dt } $$

$$\displaystyle =\frac { 1 }{ 4 } \log { \left( { t }^{ 2 }+1 \right)  } -\frac { 1 }{ 2 } \tan ^{ -1 }{ t } -\frac { 1 }{ 2 } \log { \left( t+1 \right)  } $$

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

Hence $$\displaystyle \int _{ \frac { \pi  }{ 5 }  }^{ \frac { 3\pi  }{ 10 }  } \frac { sinx }{ sinx+cosx } dx=\frac { \pi  }{ 20 } $$

$$\displaystyle \int \frac {d}{1+x^2}=\tan^{-1}x=F(x)$$
By second fundamental theorem of calculus, we obtain
$$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}=F(\sqrt 3)-F(1)$$
$$=\tan^{-1}\sqrt 3-\tan^{-1}1$$
$$\displaystyle =\frac {\pi}{3}-\frac {\pi}{4}=\frac {\pi}{12}$$

$$\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx$$

$$x=sin\theta, dx=cos\theta d\theta$$

$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta$$

$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta$$

$$\displaystyle \int _{ 0 }^{ 2\pi  }{ \frac { dx }{ 2+\sin { 2x }  }  } =\int _{ 0 }^{ 2\pi  }{ \frac { dx }{ 2+\frac { 2\tan { x }  }{ 1+\tan ^{ 2 }{ x }  }  }  } dx$$

$$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 2x }{ \frac { \sec ^{ 2 }{ x }  }{ \tan ^{ 2 }{ x } +\tan { x+1 }  } dx } $$

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

$$\displaystyle I=\frac { 1 }{ 2 } \int { \frac { 1 }{ { { t }^{ 2 }+t+1 } }  } dt.=\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( t+\frac { 1 }{ 2 }  \right)  }^{ 2 }-\frac { 3 }{ 2 }  }  } =\frac { 2\pi  }{ 3 } $$

$$I=\displaystyle \int_{\frac{1}{\sqrt{3}}}^{0}\frac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$

Substitute $$x=\displaystyle \frac{1}{z}$$
$$\Rightarrow\displaystyle  dx=-\frac{1}{z^2}dz$$

$$I=\displaystyle \int _{ \sqrt { 3 }  }^{ \infty  } \dfrac { -\dfrac { 1 }{ { z }^{ 2 } } dz }{ \left( \dfrac { 2 }{ { z }^{ 2 } } +1 \right) \sqrt { \dfrac { 1 }{ { z }^{ 2 } } +1 }  } $$

$$=\displaystyle \int _{ \sqrt { 3 }  }^{ \infty  } \dfrac { -zdz }{ \left( 2+{ z }^{ 2 } \right) \sqrt { 1+{ z }^{ 2 } }  } $$

Substitute $$\sqrt { 1+{ z }^{ 2 } } =t$$

$$\int \dfrac {dx}{x^{2}(x^{4} + 1)^{3/4}} = \int \dfrac{dx}{x^2 \cdot x^3\left(\dfrac{x^4+1}{x^4}\right)^{3/4}}=\int \dfrac {dx}{x^{5}\left (1 + \dfrac {1}{x^{4}}\right )^{3/4}}$$
Put $$1 + \dfrac {1}{x^{4}} = t\Rightarrow -\dfrac {4}{x^{5}}dx = dt$$
So, integral is
$$I = -\dfrac {1}{4}\int \dfrac {dt}{t^{3/4}} =-\dfrac{1}{4}\dfrac{(t)^{\frac{1}{4}}}{\frac{1}{4}}+c= -t^{\frac {1}{4}} + c = -\left (1 + \dfrac {1}{x^{4}}\right )^{1/4} + c$$