Integrals Test - 23

Consider, $$\displaystyle I=\int_{-\pi/4}^{\pi /4}log (cos  x +  sin  x)dx$$

$$I=\displaystyle \int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}log \left [ \sqrt{2} sin \left ( \dfrac{\pi}{4}+ x \right ) \right ]  dx$$                

let, $$\dfrac{\pi}{4}+x=t$$ $$\Rightarrow$$ $$dx=dt$$

So, $$x \rightarrow \dfrac{-\pi}{4} \Rightarrow t \rightarrow 0$$ and $$x \rightarrow \dfrac{\pi}{4} \Rightarrow t \rightarrow \dfrac{\pi}{2}$$

$$\Rightarrow$$ $$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2} \sin  t)dt$$

$$\Rightarrow$$ $$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2}) + \int_{0}^{\dfrac{\pi}{2}} \log (\sin  t)dt$$

$$\Rightarrow$$ $$\displaystyle I=\dfrac{\pi}{2}\log (\sqrt{2})-\dfrac{\pi}{2} \log (2)$$

$$\Rightarrow$$ $$\displaystyle I=\dfrac{\pi}{4}\log  2 -\dfrac{\pi}{2} \log (2)$$

$$\Rightarrow$$ $$\displaystyle I=-\dfrac{\pi}{4} \log  2$$

$$\displaystyle \int_{0}^{a}\sqrt{\dfrac{a+x}{a-x}}dx$$

$$x=a  cos  \theta =dx -a  sin  \theta  d  \theta$$

$$=\displaystyle \int_{\dfrac{\pi}{2}}^{a}\sqrt{\dfrac{a+ a  cos  \theta}{a-  a  cos  \theta}}(-a  sin  \theta)d  \theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}} \sqrt{\dfrac{1+ cos  \theta}{1- cos  \theta}}sin  \theta  d  \theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{cos \dfrac{\theta}{2}}{sin \dfrac{\theta}{2}}\cdot  2 sin \dfrac{\theta}{2} d\theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}}2  cos^2 \dfrac{\theta}{2} d \theta = a \displaystyle \int_{0}^{\dfrac{\pi}{2}}(cos \theta -1) d \theta$$

$$=a \left ( sin \theta\right )_{0}^{\frac{\pi}{2}} - (\theta)_{0}^{\frac{\pi}{2}}$$

$$=a[1-0]-\left (\dfrac{\pi}{2} \right)$$

$$=a\left (\dfrac{\pi}{2}+1 \right )$$

$$=\dfrac{a}{2}(\pi+2)$$

Let $$I=\int { \sqrt { \cfrac { x }{ 1-{ x }^{ 3 } }  } dx } $$
Substitute $$t={ x }^{ 3/2  }\Rightarrow dt=\cfrac { 3\sqrt { x }  }{ 2 } dx$$
We get
$$I=\cfrac { 2 }{ 3 } \int { \cfrac { 1 }{ \sqrt { 1-{ t }^{ 2 } }  } dt } =\cfrac { 2 }{ 3 } { sin }^{ -1 }t=\cfrac { 2 }{ 3 } { sin }^{ -1 }{ x }^{ 3/2 }$$
Therefore
$$\int _{ 0 }^{ 1 }{ Idx } =\int _{ 0 }^{ 1 }{ \cfrac { 2 }{ 3 } { sin }^{ -1 }{ x }^{ 3/2 } } =\cfrac { 2 }{ 3 } \left( \cfrac { \pi  }{ 2 } -0 \right) =\cfrac { \pi  }{ 3 } $$

Initially, we perform indefinite integration,
Let 1 + x = t
dx = dt
Putting the values back,
$$ \int (t-1){t}^{0.5} dt $$
= $$ \int {t}^{1.5} - {t}^{0.5} dt $$
= $$ \frac{2}{5} {t}^{2.5} - \frac{2}{3}{t}^{1.5} $$
Putting the value of t back,
$$ \frac{2}{5} {(1+x)}^{2.5} - \frac{2}{3}{(1+x)}^{1.5} $$
Now applying the limits to the expression,
$$ (\frac{2}{5} {(1+3)}^{2.5} - \frac{2}{3}{(1+3)}^{1.5}) - (\frac{2}{5} {(1+0)}^{2.5} - \frac{2}{3}{(1+0)}^{1.5}) $$ 
= $$ \displaystyle \frac{64}{5} - \displaystyle \frac{16}{3} - \displaystyle \frac{2}{5} + \displaystyle \frac{2}{3} $$
$$ \displaystyle \frac{62}{5} - \displaystyle \frac{14}{3} $$
= $$ \displaystyle \frac{116}{15} $$

Let $$ {x}^{0.5} = t $$
$$ \dfrac{1}{2{x}^{0.5}} dx = dt $$
Putting the values back in the given expression,
$$ \int 2 sin t dt $$
Integrating the expression, indefinitely
= $$ -2 cos t + c $$
Putting back the value of t = $$ {x}^{0.5} $$
= $$ - 2cos({x}^{0.5}) + c $$
Applying the limits,
$$ (2 cos (\dfrac{\pi}{4})) - (2 cos (\dfrac{\pi}{2})) $$
= $$ {2}^{0.5} $$

$$\int_{0}^{1} \dfrac{dx}{(x+\sqrt{x})}$$
$$x=t^2$$
$$\int_{0}^{1}\dfrac{2t  dt}{t^2+ t}=2\int_{0}^{1}\dfrac{dt}{t+1}=2 log (t+1)\int_{0}^{1}$$
$$=2  log (2)-2log (1)$$
$$=2  log  2$$

$$\int_{0}^{k}\dfrac{dx}{2+ 8x^2}=\dfrac{\pi}{16}$$
$$\dfrac{1}{2}\int_{0}^{k}\dfrac{dx}{1+(2x)^2}=\dfrac{\pi}{16}$$
$$\dfrac{1}{2}(\dfrac{1}{2})tan^{-1}(2x)\int_{0}^{k}=\dfrac{\pi}{16}$$
$$\dfrac{1}{4}tan^{-1}(2k)=\dfrac{\pi}{16}$$
$$tan^{-1}(2k)=\dfrac{\pi}{4}$$
$$2k=1$$
$$k=\dfrac{1}{2}$$

$$I_{n}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}$$

$$I_{n}+I_{n+2}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}[(cot\ \theta)^{n}+(cot\ \theta)^{n+2}]d\ \theta$$

$$=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}cosec^{2}d\ \theta$$

$$=\displaystyle \int_{^{-\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}d(cot\ \theta)$$

$$-\dfrac{(cot\ \theta)^{n+1}}{n+1}|_{^{-\pi}/_{4}}^{^{\pi}/_{2}}$$

$$-\left [ 0-\dfrac{1}{n+1} \right ]$$

$$=\dfrac{1}{n+1}$$

$$\tan^5{x}=\tan^3{x}(\sec^2{x}-1)=\tan^3{x}\sec^2{x}-\tan^3{x}$$

$$U_n=\displaystyle\int_{0}^{\pi/4} \tan^n {\theta}d\theta$$