Ex 7.11, 12 - Using properties, evaluate x dx / 1 + sin x dx - Ex 7.11

Slide27.JPG
Slide28.JPG Slide29.JPG Slide30.JPG Slide31.JPG Slide32.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
Ask Download

Transcript

Ex 7.11, 12 By using the properties of definite integrals, evaluate the integrals: ∫_0^πœ‹β–’(π‘₯ 𝑑π‘₯)/(1 + sin⁑π‘₯ ) 𝑑π‘₯ Let I=∫_0^πœ‹β–’π‘₯/(1+ sin⁑π‘₯ ) 𝑑π‘₯ ∴ I=∫_0^πœ‹β–’(πœ‹ βˆ’ π‘₯)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^πœ‹β–’( π‘₯)/(1+ sin⁑π‘₯ ) 𝑑π‘₯+∫_0^πœ‹β–’( πœ‹ βˆ’ π‘₯)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’( π‘₯ + πœ‹ βˆ’ π‘₯)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’( πœ‹)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’( 1)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ Solving I(Method 1): I=πœ‹/2 ∫_0^πœ‹β–’( 1)/(1 + sin⁑π‘₯ ) 𝑑π‘₯ Multiplying and dividing by cos⁑π‘₯ ∴, I=πœ‹/2 ∫_0^πœ‹β–’( 1/cos⁑π‘₯ )/( (1+ sin⁑π‘₯)/cos⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’sec⁑π‘₯/( 1/γ€– cos〗⁑〖π‘₯ γ€— + ( sin⁑π‘₯)/cos⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’sec⁑π‘₯/(sec⁑π‘₯ + tan⁑π‘₯ ) 𝑑π‘₯ Let sec⁑π‘₯+tan⁑〖π‘₯=𝑑〗 Differentiating both sides w.r.t. π‘₯ sec⁑π‘₯ tan⁑〖π‘₯+sec^2⁑〖π‘₯=𝑑𝑑/𝑑π‘₯γ€— γ€— sec⁑π‘₯ [tan⁑〖π‘₯+sec⁑π‘₯ γ€— ]=𝑑𝑑/𝑑π‘₯ sec⁑π‘₯ [tan⁑〖π‘₯+sec⁑π‘₯ γ€— ] 𝑑π‘₯=𝑑𝑑 sec⁑π‘₯ [𝑑] 𝑑π‘₯=𝑑𝑑 𝑑π‘₯=𝑑𝑑/(𝑑 sec⁑π‘₯ ) Putting the values of dx and (sec⁑〖x+tan⁑x γ€— )=t in(3) I=πœ‹/2 ∫_0^πœ‹β–’γ€–sec⁑π‘₯/sec⁑〖π‘₯ + tan⁑π‘₯ γ€— . 𝑑π‘₯γ€— I=πœ‹/2 ∫_0^πœ‹β–’γ€–sec⁑π‘₯/𝑑. 𝑑π‘₯γ€— I=πœ‹/2 ∫_0^πœ‹β–’γ€–sec⁑π‘₯/𝑑 ×𝑑𝑑/(𝑑 sec⁑π‘₯ )γ€— I=πœ‹/2 ∫_0^πœ‹β–’γ€–1/𝑑^2 . 𝑑𝑑〗 I=πœ‹/2 ∫_0^πœ‹β–’γ€–π‘‘^(βˆ’ 2) 𝑑𝑑〗 I= γ€–πœ‹/2 [𝑑^(βˆ’ 2 + 1)/(βˆ’ 2 + 1)]γ€—_0^πœ‹ I= γ€–πœ‹/2 [𝑑^(βˆ’1)/(βˆ’1)]γ€—_0^πœ‹ I=(βˆ’ πœ‹)/2 [1/𝑑] I=(βˆ’ πœ‹)/2 [1/sec⁑〖π‘₯ + tan⁑π‘₯ γ€— ]_0^πœ‹ I=(βˆ’ πœ‹)/2 [1/sec⁑〖(πœ‹) + tan⁑(πœ‹) γ€— βˆ’1/sec⁑〖(0) + tan⁑(0) γ€— ] I=(βˆ’ πœ‹)/2 [1/(βˆ’ 1 + 0)βˆ’ 1/( 1 + 0)] I=(βˆ’ πœ‹)/2 [βˆ’1βˆ’1] I=(βˆ’ πœ‹)/2 [βˆ’ 2] I=πœ‹ Solving I (Method 2): I=πœ‹/2 ∫_0^πœ‹β–’( 1)/(1+ sin⁑π‘₯ ) 𝑑π‘₯ Multiplying and dividing by (1βˆ’sin⁑π‘₯ ) ∴ I=πœ‹/2 ∫_0^πœ‹β–’γ€–( 1)/(1+ sin⁑π‘₯ ) Γ— (1βˆ’ sin⁑π‘₯)/(1βˆ’ sin⁑π‘₯ )γ€— . 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’(1βˆ’ sin⁑π‘₯)/(1βˆ’ sin^2⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’(1βˆ’ sin⁑π‘₯)/( γ€–cos γ€—^2⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’[1/cos^2⁑π‘₯ βˆ’sin⁑π‘₯/( γ€–cos γ€—^2⁑π‘₯ )] 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’[sec^2⁑π‘₯βˆ’sin⁑π‘₯/(cos⁑π‘₯ .γ€– cos〗⁑π‘₯ )] 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’[sec^2⁑π‘₯βˆ’tan⁑〖π‘₯ sec⁑π‘₯ γ€— ] 𝑑π‘₯ I=πœ‹/2 [[tan⁑π‘₯ ]_0^πœ‹βˆ’[sec⁑π‘₯ ]_0^πœ‹ ] I=πœ‹/2 [[π‘‘π‘Žπ‘›(πœ‹)βˆ’π‘‘π‘Žπ‘›(0)]βˆ’[𝑠𝑒𝑐(πœ‹)βˆ’π‘ π‘’π‘(0)]] I=πœ‹/2 [[0βˆ’0]βˆ’[βˆ’1βˆ’1]] I=πœ‹/2 [0βˆ’(βˆ’2)] I=πœ‹/2 [2] 𝐈=𝝅

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.