Ex 7.3, 4 - Chapter 7 Class 12 Integrals (Term 2)

Transcript

Ex 7.3, 4 sin3 (2𝑥 + 1) We know that sin⁡3𝜃=3 sin⁡𝜃−4 sin^3⁡𝜃 4 sin^3⁡𝜃=3 sin⁡𝜃−sin⁡3𝜃 sin^3⁡𝜃=(3 sin⁡𝜃 − sin⁡3𝜃)/4 Replace 𝜃 by 𝟐𝒙+𝟏 sin^3⁡(2𝑥+1)=(3 sin⁡(2𝑥 + 1) − sin⁡3(2𝑥 + 1))/4 sin^3⁡(2𝑥+1)=(3 sin⁡(2𝑥 + 1) − sin⁡(6𝑥 + 3))/4 Thus, our equation becomes . ∫1▒〖sin3 (2𝑥+1) 〗 𝑑𝑥 =∫1▒(3 sin⁡(2𝑥 + 1) − sin⁡(6𝑥 + 3))/4 𝑑𝑥 =1/4 ∫1▒(3 sin⁡(2𝑥+1)−sin⁡(6𝑥+3) ) 𝑑𝑥 =1/4 [3∫1▒sin⁡(2𝑥+1) 𝑑𝑥−∫1▒sin⁡(6𝑥+3) 𝑑𝑥] =1/4 [3⤶×1/2 (−cos⁡(2𝑥+1) )−1/6 (−cos⁡(6𝑥+3)+𝐶)" " ] =1/4 [(−3)/2 cos⁡(2𝑥+1)+1/6 cos⁡(6𝑥+3) ]+𝐶 =(−3)/8 cos⁡(2𝑥+1)+1/24 𝒄𝒐𝒔⁡𝟑(𝟐𝒙+𝟏)+𝐶 ∫1▒sin⁡(𝑎𝑥+𝑏) 𝑑𝑥=−〖𝑐𝑜𝑠 〗⁡(𝑎𝑥 + 𝑏)/𝑎 +𝐶 We know that 𝑐𝑜𝑠⁡3𝜃=4 〖𝑐𝑜𝑠〗^3⁡𝜃−3 𝑐𝑜𝑠⁡𝜃 Replace 𝜃 by 2𝑥+1 𝑐𝑜𝑠⁡3(2𝑥+1)=4 〖𝑐𝑜𝑠〗^3⁡(2𝑥+1)−3 𝑐𝑜𝑠⁡(2𝑥+1) =(−3)/8 cos⁡(2𝑥+1)+1/24 [𝟒 〖𝒄𝒐𝒔〗^𝟑⁡(𝟐𝒙+𝟏)−𝟑 𝒄𝒐𝒔⁡(𝟐𝒙+𝟏) ]+𝐶 =(−3)/8 cos⁡(2𝑥+1)+4/24 cos^3⁡(2𝑥+1)−3/24 cos⁡(2𝑥+1)+𝐶 =(−3)/8 cos⁡(2𝑥+1)+1/6 cos^3⁡(2𝑥+1)−1/8 cos⁡(2𝑥+1)+𝐶 =(−3)/8 cos⁡(2𝑥+1)−1/8 cos⁡(2𝑥+1)+1/6 cos^3⁡(2𝑥+1)+𝐶 =(−4)/8 cos⁡(2𝑥+1)+1/6 cos^3⁡(2𝑥+1)+𝐶 =(−𝟏)/𝟐 𝒄𝒐𝒔⁡(𝟐𝒙+𝟏)+𝟏/𝟔 〖𝒄𝒐𝒔〗^𝟑⁡(𝟐𝒙+𝟏)+𝑪