Integrals Test - 51

Consider $$\displaystyle I_{t+2}+I_{t}-2I_{t+1}$$
$$\displaystyle \int_{0}^{\pi /2}\frac{\left ( \sin ^{2}tx+\sin^{2}\left ( t+2 \right )x-2\sin ^{2}\left ( t+1 \right )x \right )dx}{\sin ^{2}x}$$
$$\displaystyle \int_{0}^{\pi /2}\frac{\left ( \sin ^{2}\left ( t+2 \right )x-\sin^{2}\left ( t+1 \right )x-\sin ^{2}\left ( t+1 \right )x+\sin^{2}tx \right )dx}{\sin ^{2}x}$$
$$\displaystyle \int_{0}^{\pi /2}\frac{\left [ \sin \left ( 2t+3 \right )x-\sin \left ( 2t+1 \right )x \right ]dx}{\sin x}$$
$$\displaystyle

=2\int_{0}^{\pi /2}\cos \left ( 2t+2 \right )x dx=2\left ( \frac{\sin

\left ( 2t+2 \right )x}{2\left ( t+1 \right )} \right )^{\pi /2}_{0}$$
$$\displaystyle = \frac{1}{\left ( t+1 \right )}\left ( 0 \right )=0$$
$$\displaystyle \therefore I_{t},I_{t+1},I_{t+2}\epsilon A.P.$$
Short Cut Method : 
Substitute $$t=1, 2, 3$$ in given integral we get
$$\displaystyle I_{1}= \frac{\pi}{2},I_{2}=\pi , I_{3}=\frac{3\pi }{2} $$
$$\displaystyle I_{1},I_{2}, I_{3}\epsilon A.P.$$

$$\displaystyle F\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt+\int_{1}^{1/x}\frac{\ln t}{1+t}dt$$
$$\displaystyle F\left ( x \right )=\int_{1}^{x}\left ( \frac{\ln t}{1+t}+\frac{\ln t}{\left ( 1+t \right )t} \right )dt=\int_{1}^{t}\frac{\ln t}{t}dt=\frac{1}{2}\left ( \ln x \right )^{2}$$
$$F\left ( e \right )=1/2$$

Let $$\displaystyle I=\int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } }{ \frac { \cos { x } }{ \cos { x } +\sin { x } } } dx$$

Multiply numerator and denominator by $$sec^{ 3 }{ \left( x \right) }$$
$$\displaystyle I=\int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } }{ \frac { sec^{ 2 }{ x } }{ sec^{ 2 }{ x }+sec^{ 2 }{ x }\tan { x } } } dx=\int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } }{ \frac { sec^{ 2 }{ x } }{ 1+\tan { x } +\tan ^{ 2 }{ x } +\tan ^{ 3 }{ x } } } dx$$

Put $$t=\tan { x } \Rightarrow dt=sec^{ 2 }{ x }dx$$
$$\displaystyle \therefore I=\int _{ \tan { \frac { \pi }{ 5 } } }^{ \tan { \frac { 3\pi }{ 10 } } }{ \frac { 1 }{ { u }^{ 2 }+{ u }^{ 3 }+u+1 } } du=\int _{ \tan { \frac { \pi }{ 5 } } }^{ \frac { 3\pi }{ 10 } }{ \left( \frac { 1-u }{ 2\left( { u }^{ 2 }+1 \right) } +\frac { 1 }{ 2\left( u+1 \right) } \right) du } $$

$$\displaystyle =\frac { 1 }{ 2 } \int _{ \tan { \frac { \pi }{ 5 } } }^{ \tan { \frac { 3\pi }{ 10 } } }{ \left( \frac { 1 }{ { u }^{ 2 }+1 } -\frac { u }{ { u }^{ 2 }+1 } \right) } du+\frac { 1 }{ 2 } \int _{ \tan { \frac { \pi }{ 5 } } }^{ \tan { \frac { 3\pi }{ 10 } } }{ u+1 } du$$

$$\displaystyle =\left[ -\frac { 1 }{ 4 } \log { \left( { u }^{ 2 }+1 \right) +\frac { 1 }{ 2 } } \log { \left( u+1 \right) +\frac { 1 }{ 2 } \tan ^{ -1 }{ u } } \right] _{ \log { \frac { \pi }{ 5 } } }^{ \log { \frac { 3\pi }{ 10 } } }=\frac { \pi }{ 20 } $$

Let $$I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta  }{ 5+3\cos { \theta  }  }  } =\int _{ 0 }^{ \pi /2 }{ \frac { d\theta  }{ 5+3.\frac { 1-\tan ^{ 2 }{ \frac { \theta  }{ 2 }  }  }{ 1+\tan ^{ 2 }{ \frac { \theta  }{ 2 }  }  }  }  } $$

$$\displaystyle I= \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx$$
$$\displaystyle = \frac{1}{2}\int_{0}^{\pi /2}\sin x\log \sin ^{2}x dx$$
$$\displaystyle = \frac{1}{2}\int_{0}^{\pi /2}\sin x\log \left ( 1-\cos ^{2}x \right )dx$$
$$\displaystyle = \frac{1}{2}\int_{0}^{1}\log \left ( 1-t^{2} \right )dt$$ where $$\displaystyle t= \cos x$$
$$\displaystyle = \frac{1}{2}\int_{0}^{1}\left [ -t^{2}-\frac{\left ( t^{2} \right )^{2}}{2}-\frac{\left ( t^{2} \right )^{3}}{3}-\cdots  \right ]dt$$
$$\displaystyle = -\left [ \frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\cdots  \right ]$$
$$\displaystyle = -\left [ \left ( \frac{1}{2}-\frac{1}{3} \right )+\left ( \frac{1}{4}-\frac{1}{5} \right )+\left ( \frac{1}{6}-\frac{1}{7} \right )+\cdots  \right ]$$
$$\displaystyle = -1+\left ( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots  \right )$$.
$$\displaystyle = -\log _{e}e+\log _{e}\left ( 1+1 \right )= \log _{e}\left ( 2/e \right )$$

Substitute $$\displaystyle x=\tan \theta \therefore I_2=\int_{0}^{\pi /4}\frac{\theta }{\tan \theta }.\sec ^{2}\theta d\theta $$ or $$\displaystyle I_{2}=\int_{0}^{\pi /4}\frac{2\theta }{2\sin \theta \cos \theta }d\theta =\int_{0}^{\pi /4} \frac{2\theta }{\sin 2\theta }d\theta $$
Now substitute $$\displaystyle 2\theta =t \therefore I_{2}=\frac{1}{2}\int_{0}^{\pi /2}\frac{t}{\sin t}dt=\frac{1}{2}I_{1} $$ $$\displaystyle \therefore \frac{I_{1}}{I_{2}}=2\Rightarrow \left ( C \right )$$  

$$\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ t }^{ 2 } } } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt$$
$$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right] $$
$$\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 } $$

$$\displaystyle I=\int _{ 0 }^{ a }{ { x }^{ 4 }\sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } =\int _{ 0 }^{ a }{ { x }^{ 5 }\sqrt { { \left( \frac { a }{ x }  \right)  }^{ 2 }-1 } dx } $$

Put $$\displaystyle { \left( \frac { a }{ x }  \right)  }^{ 2 }=t\Rightarrow \left( -\frac { 2{ a }^{ 2 } }{ { x }^{ 3 } }  \right) dx=dt$$

$$\displaystyle I=-\int _{ \infty  }^{ 1 }{ \frac { { a }^{ 8 } }{ { t }^{ 4 } } \sqrt { { t }^{ 2 }-1 } dt } =\int _{ 1 }^{ \infty  }{ \frac { { a }^{ 8  }\sqrt { { t }^{ 2 }-1 }  }{ \sqrt { { t }^{ 2 } }  } dt } $$

$$\displaystyle ={ a }^{ 8  }\int _{ 1 }^{ \infty  }{ \sqrt { 1-\frac { 1 }{ { t }^{ 2 } }  } dt } =\frac { \pi { a }^{ 6 } }{ 32 } $$

In $$\displaystyle N^{r}$$ put $$\displaystyle x=\sin \theta ,$$ then $$\displaystyle \sin ^{-1}\left ( 1-2\sin ^{2}\theta  \right )+\cos ^{-1}\left ( 2\sin \theta \cos \theta  \right )=\sin ^{-1}\left ( \cos 2\theta  \right )+\cos ^{-1}\left ( \sin 2\theta  \right )$$ $$\displaystyle =\frac{\pi }{2}-\cos ^{-1}\left ( \cos 2\theta  \right )+\frac{\pi }{2}-\sin ^{-1}\left ( \sin 2\theta  \right )$$ $$\displaystyle =\pi -2\theta -2\theta =\pi -4\sin ^{-1}x$$
 $$\displaystyle \therefore I=\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\left [ \pi -4\sin -\mid x \right ]dx$$ $$\displaystyle =2 \int_{0}^{\frac{1}{\sqrt{2}}}\pi \frac{x^{8}}{1-x^{4}}dx+0$$ by Prop.V $$\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}\frac{x^{8}-1+1}{1-x^{4}}dx$$ $$\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}\left [ -\left ( x^{4}+1 \right )+\frac{1}{\left ( 1-x^{2} \right )\left ( 1+x^{2} \right )} \right ]dx$$ $$\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}-\left ( x^{4}+1 \right )+\frac{1}{2}\left \{ \frac{1}{1-x^{2}}+\frac{1}{1+x^{2}} \right \}dx$$ $$\displaystyle =2x\left [ -\left ( \frac{x^{5}}{5}+x \right )+\frac{1}{4}\log \frac{1+x}{1-x} +\frac{1}{2}\tan -1x\right ]^{1/\sqrt{2}}_{0}$$ $$\displaystyle =\pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$$ 

Ans: A

Given $$\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}$$       ....(1)

Consider, $$I= \int _{ 0 }^{ \pi /3 } \dfrac { \cos  \theta  }{ 3+4\sin  \theta  } d\theta $$
Substitute $$\sin  \theta =t$$
$$\Rightarrow \cos  \theta d\theta =dt$$

$$I=\displaystyle \int _{ 0 }^{ \sqrt { 3 } /2 } \frac { dt }{ 3+4t } $$
Substitute $$3+4t=u$$
$$4dt=du$$

$$\therefore I=\displaystyle \frac {1}{4}\int _{ 3 }^{ 3+2\sqrt { 3 } } \frac { du }{ u } $$

$$\displaystyle =\frac{1}{4}[\log { u } ]_{ 3 }^{3+ 2\sqrt { 3 } }$$

$$\displaystyle I=\frac{1}{4} [\log { (3+2\sqrt { 3 } )-\log { 3 }  }] $$

$$\Rightarrow I=\displaystyle \frac{1}{4}\log  \frac { 3+2\sqrt { 3 }  }{ 3 } $$
So, on comparing with (1), we get
$$\lambda =\displaystyle \frac{1}{4}$$