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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⁡𝑥 sin⁡(𝑥 + 𝛼) ) Solving sin^3⁡𝑥 sin⁡(𝑥 + 𝛼) =sin^3⁡𝑥 [sin⁡𝑥 cos⁡𝛼+cos⁡𝑥.sin⁡𝛼 ] =〖sin^4 𝑥〗⁡cos⁡𝛼 +cos⁡𝑥.sin^3⁡𝑥 sin⁡𝛼 =〖sin^4 𝑥〗⁡cos⁡𝛼 +cos⁡𝑥.sin^3⁡𝑥 sin⁡𝛼×sin⁡𝑥/sin⁡𝑥 =sin^4⁡𝑥 [cos⁡𝛼+cos⁡𝑥 . sin⁡𝛼.1/sin⁡𝑥 ] Hence Using 𝑠𝑖𝑛⁡(𝐴+𝐵)=𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵+𝑐𝑜𝑠⁡𝐴.𝑠𝑖𝑛⁡𝐵 =sin^4⁡𝑥 [cos⁡𝛼+cos⁡𝑥/sin⁡𝑥 . sin⁡𝛼 ] =sin^4⁡𝑥 [cos⁡𝛼+cot⁡𝑥 sin⁡𝛼 ] Therefore sin^3⁡𝑥 sin⁡(𝑥+𝛼)=sin^4⁡𝑥 (cos⁡𝛼+cot⁡𝑥.sin⁡𝛼 ) Now ∫1▒1/√(sin^3⁡𝑥 sin⁡(𝑥 + 𝛼) ) 𝑑𝑥 =∫1▒1/√(sin^4⁡𝑥 (cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 ) ) 𝑑𝑥 =∫1▒〖1/√(sin^4⁡𝑥 )×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2⁡𝑥 ×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 Let cos⁡𝛼+cot⁡𝑥. sin⁡𝛼=𝑡 Diff w.r.t. x 𝑑(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos⁡𝛼 )/𝑑𝑥+sin⁡𝛼 𝑑(cot⁡𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 =∫1▒〖1/√(sin^4⁡𝑥 )×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2⁡𝑥 ×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 Let cos⁡𝛼+cot⁡𝑥. sin⁡𝛼=𝑡 Diff w.r.t. x 𝑑(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos⁡𝛼 )/𝑑𝑥+sin⁡𝛼 𝑑(cot⁡𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 0+sin⁡𝛼 (−𝑐𝑜𝑠𝑒𝑐^2 𝑥)=𝑑𝑡/𝑑𝑥 −sin⁡𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−sin⁡𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥) 𝑑𝑥=1/(−sin⁡𝛼 ) . 1/(𝑐𝑜𝑠𝑒𝑐^2 𝑥) . 𝑑𝑡 𝑑𝑥=1/(−sin⁡𝛼 ) . sin^2⁡𝑥. 𝑑𝑡 Now our equation becomes ∫1▒1/(sin^2⁡𝑥 √(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 ) ) 𝑑𝑥 =∫1▒1/(sin^2⁡𝑥 √𝑡 )×1/(−sin⁡𝛼 )×sin^2⁡𝑥 𝑑𝑡 cos⁡𝛼 &sin⁡𝛼 "are constant" █(" " @"&" 𝑑(cot⁡𝑥 )/𝑑𝑥=−𝑐𝑜𝑠𝑒𝑐 𝑥) =1/(−sin⁡𝛼 ) ∫1▒1/√𝑡 𝑑𝑡 =(−1)/sin⁡𝛼 ∫1▒(𝑡)^((−1)/2) 𝑑𝑡 =(−1)/sin⁡𝛼 [𝑡^((−1)/2 + 1)/((−1)/2 + 1) +𝐶] =(−1)/sin⁡𝛼 [𝑡^(1/2)/(1/2) +𝐶] =(−𝟏)/𝒔𝒊𝒏⁡𝜶 [𝟐√𝒕 +𝑪] Putting back value of 𝑡=√(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) =(−1)/sin⁡𝛼 [2√(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 )+𝐶] =(−2)/sin⁡𝛼 √(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) −1/sin⁡𝛼 . 𝐶 =(−2)/sin⁡𝛼 √(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) +𝐶 Now, From (1) sin^3⁡𝑥 sin⁡(𝑥+𝛼)=sin^4⁡𝑥 (𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) (sin^3⁡𝑥 sin⁡(𝑥 + 𝛼))/sin^4⁡𝑥 =𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 〖𝑠𝑖𝑛 〗⁡(𝑥 + 𝛼)/sin⁡𝑥 =𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 Thus, Answer =(−𝟐)/𝒔𝒊𝒏⁡𝒙 √(𝒔𝒊𝒏⁡(𝒙 + 𝜶)/𝐬𝐢𝐧⁡𝒙 ) + 𝑪

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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.