Mathematics

Let $${I_n} = \int_0^\pi  {\dfrac{{{{\sin }^2}\left( {nx} \right)}}{{{{\sin }^2}x}}} \,dx\,\,,\,n \in N$$, then


ANSWER

$${I_{n + 2}} + {I_n} = {21_{n + 1}}$$


SOLUTION
$${ I }_{ n }=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin ^{ 2 }{ nx }  }{ \sin ^{ 2 }{ x }  } dx } ;\quad { I }_{ n+1 }=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin ^{ 2 }{ \left( n+1 \right)  }  }{ \sin ^{ 2 }{ x }  } dx } $$

$${ I }_{ n+1 }-{ I }_{ n }=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin ^{ 2 }{ \left( n+1 \right) x- } \sin ^{ 2 }{ nx }  }{ \sin ^{ 2 }{ x }  }  } $$

$$=\displaystyle\int _{ 0 }^{ \pi  }{ \frac { \sin ^{ 2 }{ \left( 2n+1 \right) x. } \sin { x }  }{ \sin ^{ 2 }{ x }  }  } dx$$

$$=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin ^{ 2 }{ \left( 2n+1 \right) x }  }{ \sin { x }  }  } dx$$

$$I_{n+2}-I_{n+1}=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin { \left( 2n+3 \right) x }  }{ \sin { x }  }  } dx$$

$$I_{n+2}-2I_{n+1}+I_n=\displaystyle \int _{ 0 }^{ \pi  }{ \frac { \sin { \left( 2n+3 \right) x- } \sin { \left( 2n+1 \right) x }  }{ \sin { x }  }  } dx$$

$$=\displaystyle\int _{ 0 }^{ \pi  }{ \frac { \cos { \left( 2n+3 \right) x- } \sin { x }  }{ \sin { x }  }  } dx$$

$$=\displaystyle\int _{ 0 }^{ \pi  }{ 2\cos 2(n+1)x dx}$$

$$=\left. 2\dfrac{\sin (2n+2)x}{2x+2}\right|_{0}^{\pi}=0$$

so we have
$$I_{n+2}+I_n=2I_{n+1}$$
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Subjective Hard
Solve $$\displaystyle\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 2 }-5x+6 }  } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
$$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{ 1 + \sin \, x}$$ =
  • A. $$\dfrac{\pi}{6} $$
  • B. $$\dfrac{\pi}{3}$$
  • C. Cannot be valued
  • D. $$\pi$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
The value of $$\displaystyle \int \sqrt{2}\left ( \frac{\sin x}{\sin \left ( x-\dfrac {\pi }4 \right )} \right )dx$$ is
  • A. $$\displaystyle x-\log \left | \sin \left ( x-\frac{\pi }{4} \right ) \right |+c$$
  • B. $$\displaystyle x+\log \left | \cos \left ( x-\frac{\pi }{4} \right ) \right |+c$$
  • C. $$\displaystyle x-\log \left | \cos \left ( x-\frac{\pi }{4} \right ) \right |+c$$
  • D. $$\displaystyle x+\log \left | \sin \left ( x-\frac{\pi }{4} \right ) \right |+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Medium
Solve $$\displaystyle \int {\tan x\,\ln \left( {\cos x} \right)} \,dx$$
  • A. $$\dfrac {\ln^2(\cos x)}{2}+C$$
  • B. $$\dfrac {\ln^2(\sin x)}{2}+C$$
  • C. $$\dfrac {\ln(\cos x)}{2}+C$$
  • D. $$-\dfrac {(\ln(\cos x))^2}{2}+C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Single Correct Medium
$$\displaystyle \int_2^3\dfrac{dx}{x^2-1}$$
  • A. $$log\dfrac{3}{2}$$
  • B. $$2log\dfrac{3}{2}$$
  • C. $$\log\dfrac{3}{2}$$
  • D. $$\dfrac{1}{2}log\dfrac{3}{2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer