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


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Misc 18 Integrate the function 1/√(sin^3⁑π‘₯ sin⁑(π‘₯ + 𝛼) ) Solving sin^3⁑π‘₯ sin⁑(π‘₯ + 𝛼) =sin^3⁑π‘₯ [sin⁑π‘₯ cos⁑𝛼+cos⁑π‘₯.sin⁑𝛼 ] =γ€–sin^4 π‘₯〗⁑cos⁑𝛼 +cos⁑π‘₯.sin^3⁑π‘₯ sin⁑𝛼 =γ€–sin^4 π‘₯〗⁑cos⁑𝛼 +cos⁑π‘₯.sin^3⁑π‘₯ sin⁑𝛼×sin⁑π‘₯/sin⁑π‘₯ =sin^4⁑π‘₯ [cos⁑𝛼+cos⁑π‘₯ . sin⁑𝛼.1/sin⁑π‘₯ ] Hence Using 𝑠𝑖𝑛⁑(𝐴+𝐡)=𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅+π‘π‘œπ‘ β‘π΄.𝑠𝑖𝑛⁑𝐡 =sin^4⁑π‘₯ [cos⁑𝛼+cos⁑π‘₯/sin⁑π‘₯ . sin⁑𝛼 ] =sin^4⁑π‘₯ [cos⁑𝛼+cot⁑π‘₯ sin⁑𝛼 ] Therefore sin^3⁑π‘₯ sin⁑(π‘₯+𝛼)=sin^4⁑π‘₯ (cos⁑𝛼+cot⁑π‘₯.sin⁑𝛼 ) Now ∫1β–’1/√(sin^3⁑π‘₯ sin⁑(π‘₯ + 𝛼) ) 𝑑π‘₯ =∫1β–’1/√(sin^4⁑π‘₯ (cos⁑𝛼 + cot⁑π‘₯ . sin⁑𝛼 ) ) 𝑑π‘₯ =∫1β–’γ€–1/√(sin^4⁑π‘₯ )Γ—1/√(cos⁑𝛼 + cot⁑π‘₯ . sin⁑𝛼 )γ€— 𝑑π‘₯ =∫1β–’γ€–1/sin^2⁑π‘₯ Γ—1/√(cos⁑𝛼 + cot⁑π‘₯ . sin⁑𝛼 )γ€— 𝑑π‘₯ Let cos⁑𝛼+cot⁑π‘₯. sin⁑𝛼=𝑑 Diff w.r.t. x 𝑑(cos⁑𝛼 + cot⁑π‘₯ sin⁑𝛼 )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑(cos⁑𝛼 )/𝑑π‘₯+sin⁑𝛼 𝑑(cot⁑π‘₯ )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ =∫1β–’γ€–1/√(sin^4⁑π‘₯ )Γ—1/√(cos⁑𝛼 + cot⁑π‘₯ . sin⁑𝛼 )γ€— 𝑑π‘₯ =∫1β–’γ€–1/sin^2⁑π‘₯ Γ—1/√(cos⁑𝛼 + cot⁑π‘₯ . sin⁑𝛼 )γ€— 𝑑π‘₯ Let cos⁑𝛼+cot⁑π‘₯. sin⁑𝛼=𝑑 Diff w.r.t. x 𝑑(cos⁑𝛼 + cot⁑π‘₯ sin⁑𝛼 )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑(cos⁑𝛼 )/𝑑π‘₯+sin⁑𝛼 𝑑(cot⁑π‘₯ )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ 0+sin⁑𝛼 (βˆ’π‘π‘œπ‘ π‘’π‘^2 π‘₯)=𝑑𝑑/𝑑π‘₯ βˆ’sin⁑𝛼 π‘π‘œπ‘ π‘’π‘^2 π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’sin⁑𝛼 π‘π‘œπ‘ π‘’π‘^2 π‘₯) 𝑑π‘₯=1/(βˆ’sin⁑𝛼 ) . 1/(π‘π‘œπ‘ π‘’π‘^2 π‘₯) . 𝑑𝑑 𝑑π‘₯=1/(βˆ’sin⁑𝛼 ) . sin^2⁑π‘₯. 𝑑𝑑 Now our equation becomes ∫1β–’1/(sin^2⁑π‘₯ √(cos⁑𝛼 + cot⁑π‘₯ sin⁑𝛼 ) ) 𝑑π‘₯ =∫1β–’1/(sin^2⁑π‘₯ βˆšπ‘‘ )Γ—1/(βˆ’sin⁑𝛼 )Γ—sin^2⁑π‘₯ 𝑑𝑑 cos⁑𝛼 &sin⁑𝛼 "are constant" β–ˆ(" " @"&" 𝑑(cot⁑π‘₯ )/𝑑π‘₯=βˆ’π‘π‘œπ‘ π‘’π‘ π‘₯) =1/(βˆ’sin⁑𝛼 ) ∫1β–’1/βˆšπ‘‘ 𝑑𝑑 =(βˆ’1)/sin⁑𝛼 ∫1β–’(𝑑)^((βˆ’1)/2) 𝑑𝑑 =(βˆ’1)/sin⁑𝛼 [𝑑^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) +𝐢] =(βˆ’1)/sin⁑𝛼 [𝑑^(1/2)/(1/2) +𝐢] =(βˆ’πŸ)/π’”π’Šπ’β‘πœΆ [πŸβˆšπ’• +π‘ͺ] Putting back value of 𝑑=√(π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 ) =(βˆ’1)/sin⁑𝛼 [2√(π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 )+𝐢] =(βˆ’2)/sin⁑𝛼 √(π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 ) βˆ’1/sin⁑𝛼 . 𝐢 =(βˆ’2)/sin⁑𝛼 √(π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 ) +𝐢 Now, From (1) sin^3⁑π‘₯ sin⁑(π‘₯+𝛼)=sin^4⁑π‘₯ (π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 ) (sin^3⁑π‘₯ sin⁑(π‘₯ + 𝛼))/sin^4⁑π‘₯ =π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 〖𝑠𝑖𝑛 〗⁑(π‘₯ + 𝛼)/sin⁑π‘₯ =π‘π‘œπ‘ β‘π›Ό+π‘π‘œπ‘‘β‘π‘₯. 𝑠𝑖𝑛⁑𝛼 Thus, Answer =(βˆ’πŸ)/π’”π’Šπ’β‘π’™ √(π’”π’Šπ’β‘(𝒙 + 𝜢)/𝐬𝐒𝐧⁑𝒙 ) + π‘ͺ

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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.