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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Ex 7.10, 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‘₯ Multiplying and dividing by (1βˆ’sin⁑π‘₯ ) I=πœ‹/2 ∫_0^πœ‹β–’γ€–( 1)/(1 + sin⁑π‘₯ ) Γ— (1 βˆ’ sin⁑π‘₯)/(1 βˆ’ sin⁑π‘₯ )γ€— . 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’(1 βˆ’ sin⁑π‘₯)/(1 βˆ’ sin^2⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’(1 βˆ’ 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.