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Last updated at May 29, 2023 by Teachoo
Misc 18 Integrate the function 1/√(sin^3𝑥 sin(𝑥 + 𝛼) ) Solving sin^3𝑥 sin(𝑥 + 𝛼) =sin^3𝑥 [sin𝑥 cos𝛼+cos𝑥.sin𝛼 ] =〖sin^4 𝑥〗cos𝛼 +cos𝑥.sin^3𝑥 sin𝛼 =〖sin^4 𝑥〗cos𝛼 +cos𝑥.sin^3𝑥 sin𝛼×sin𝑥/sin𝑥 =sin^4𝑥 [cos𝛼+cos𝑥 . sin𝛼.1/sin𝑥 ] Hence Using 𝑠𝑖𝑛(𝐴+𝐵)=𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴.𝑠𝑖𝑛𝐵 =sin^4𝑥 [cos𝛼+cos𝑥/sin𝑥 . sin𝛼 ] =sin^4𝑥 [cos𝛼+cot𝑥 sin𝛼 ] Therefore sin^3𝑥 sin(𝑥+𝛼)=sin^4𝑥 (cos𝛼+cot𝑥.sin𝛼 ) Now ∫1▒1/√(sin^3𝑥 sin(𝑥 + 𝛼) ) 𝑑𝑥 =∫1▒1/√(sin^4𝑥 (cos𝛼 + cot𝑥 . sin𝛼 ) ) 𝑑𝑥 =∫1▒〖1/√(sin^4𝑥 )×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2𝑥 ×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 Let cos𝛼+cot𝑥. sin𝛼=𝑡 Diff w.r.t. x 𝑑(cos𝛼 + cot𝑥 sin𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos𝛼 )/𝑑𝑥+sin𝛼 𝑑(cot𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 =∫1▒〖1/√(sin^4𝑥 )×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2𝑥 ×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 Let cos𝛼+cot𝑥. sin𝛼=𝑡 Diff w.r.t. x 𝑑(cos𝛼 + cot𝑥 sin𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos𝛼 )/𝑑𝑥+sin𝛼 𝑑(cot𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 0+sin𝛼 (−𝑐𝑜𝑠𝑒𝑐^2 𝑥)=𝑑𝑡/𝑑𝑥 −sin𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−sin𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥) 𝑑𝑥=1/(−sin𝛼 ) . 1/(𝑐𝑜𝑠𝑒𝑐^2 𝑥) . 𝑑𝑡 𝑑𝑥=1/(−sin𝛼 ) . sin^2𝑥. 𝑑𝑡 Now our equation becomes ∫1▒1/(sin^2𝑥 √(cos𝛼 + cot𝑥 sin𝛼 ) ) 𝑑𝑥 =∫1▒1/(sin^2𝑥 √𝑡 )×1/(−sin𝛼 )×sin^2𝑥 𝑑𝑡 cos𝛼 &sin𝛼 "are constant" █(" " @"&" 𝑑(cot𝑥 )/𝑑𝑥=−𝑐𝑜𝑠𝑒𝑐 𝑥) =1/(−sin𝛼 ) ∫1▒1/√𝑡 𝑑𝑡 =(−1)/sin𝛼 ∫1▒(𝑡)^((−1)/2) 𝑑𝑡 =(−1)/sin𝛼 [𝑡^((−1)/2 + 1)/((−1)/2 + 1) +𝐶] =(−1)/sin𝛼 [𝑡^(1/2)/(1/2) +𝐶] =(−𝟏)/𝒔𝒊𝒏𝜶 [𝟐√𝒕 +𝑪] Putting back value of 𝑡=√(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) =(−1)/sin𝛼 [2√(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 )+𝐶] =(−2)/sin𝛼 √(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) −1/sin𝛼 . 𝐶 =(−2)/sin𝛼 √(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) +𝐶 Now, From (1) sin^3𝑥 sin(𝑥+𝛼)=sin^4𝑥 (𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) (sin^3𝑥 sin(𝑥 + 𝛼))/sin^4𝑥 =𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 〖𝑠𝑖𝑛 〗(𝑥 + 𝛼)/sin𝑥 =𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 Thus, Answer =(−𝟐)/𝒔𝒊𝒏𝒙 √(𝒔𝒊𝒏(𝒙 + 𝜶)/𝐬𝐢𝐧𝒙 ) + 𝑪