Example 7 - Chapter 7 Class 12 Integrals

Transcript

Example 7 Find (i) ∫1▒cos^2⁡𝑥 𝑑𝑥 ∫1▒cos^2⁡𝑥 𝑑𝑥 =∫1▒((cos⁡2𝑥 + 1)/2) 𝑑𝑥 = 1/2 ∫1▒(cos⁡2𝑥+1) 𝑑𝑥 = 1/2 [∫1▒cos⁡2𝑥 𝑑𝑥+∫1▒1 𝑑𝑥] We know cos⁡2𝑥=2 cos^2⁡𝑥−1 cos⁡2𝑥+1=2 cos^2⁡𝑥 (cos⁡2𝑥 + 1)/2=cos^2⁡𝑥 ∫1▒𝒄𝒐𝒔⁡𝟐𝒙 𝒅𝒙 Let 2𝑥=𝑡 2 =𝑑𝑡/𝑑𝑥 𝑑𝑥=1/2 𝑑𝑡 ∫1▒cos⁡𝑡 . 1/2 𝑑𝑡 =1/2 (sin⁡𝑡+𝐶1) Putting value of 𝑡 = 2𝑥 =1/2 sin⁡2𝑥+𝐶1 ∫1▒𝟏 𝒅𝒙 =∫1▒𝑥^0 𝑑𝑥 =[𝑥^(0 + 1)/(0 + 1)]+𝐶 =𝑥+𝐶2 Thus, ∫1▒cos^2⁡𝑥 𝑑𝑥=1/2 [∫1▒cos⁡2𝑥 𝑑𝑥+∫1▒1 𝑑𝑥] =1/2 [1/2 sin⁡2𝑥+𝐶1+𝑥+𝐶2] =1/4 sin⁡2𝑥+𝑥/2+1/2(𝐶1+𝐶2) =𝒙/𝟐+𝟏/𝟒 𝐬𝐢𝐧⁡𝟐𝒙+𝑪 ("From (1) and (2) " ) ("Let" 𝐶1+𝐶2=𝐶) Example 7 Find (ii) ∫1▒〖sin⁡〖2𝑥 〗 cos⁡3𝑥 〗 𝑑𝑥 We know that 2 sin⁡𝐴 cos⁡𝐵=sin⁡(𝐴+𝐵)+sin⁡(𝐴−𝐵) sin⁡𝐴 cos⁡𝐵=1/2 [sin⁡(𝐴+𝐵)+sin⁡(𝐴−𝐵) ] Replace A by 2𝑥 & B by 3𝑥 sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡(2𝑥+3𝑥)+sin⁡(2𝑥−3𝑥) ] sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡(5𝑥)+sin⁡(−𝑥) ] sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡5𝑥−sin⁡𝑥 ] ∫1▒(sin⁡〖2𝑥 〗 cos⁡3𝑥 ) 𝑑𝑥=1/2 ∫1▒(sin⁡5𝑥−sin⁡𝑥 ) 𝑑𝑥 =1/2 [∫1▒sin⁡5𝑥 𝑑𝑥−∫1▒sin⁡𝑥 𝑑𝑥] ∫1▒𝒔𝒊𝒏⁡𝟓𝒙 𝒅𝒙 Let 5𝑥=𝑡 5 = 𝑑𝑡/𝑑𝑥 𝑑𝑥=1/5 𝑑𝑡 =∫1▒sin⁡𝑡 . 1/5 𝑑𝑡 =1/5 ∫1▒sin⁡𝑡 . 𝑑𝑡 =1/5 (〖−cos〗⁡𝑡+𝐶1) =−1/5 cos⁡𝑡+1/5 𝐶1 Putting value of 𝑡 =−1/5 cos⁡5𝑥+1/5 𝐶1 ∫1▒sin⁡𝑥 𝑑𝑥 =−cos⁡𝑥+𝐶2 Thus, ∫1▒〖sin⁡〖2𝑥 〗 cos⁡3𝑥 〗=1/2 [∫1▒〖sin⁡5𝑥 𝑑𝑥〗−∫1▒sin⁡𝑥 𝑑𝑥] =1/2 [−1/5 cos⁡5𝑥+1/5 𝐶1−(−cos⁡𝑥 )+𝐶2] =1/2 [−1/5 cos⁡5𝑥+cos⁡𝑥+1/5 𝐶1−𝐶2] =1/2 [−1/5 cos⁡5𝑥+cos⁡𝑥+𝐶] =(−𝟏)/𝟏𝟎 𝐜𝐨𝐬⁡𝟓𝒙+𝟏/𝟐 𝐜𝐨𝐬⁡𝒙+𝑪 Example 7 Find (iii) ∫1▒sin^3⁡𝑥 𝑑𝑥 We know that 𝑠𝑖𝑛 3𝑥=3 𝑠𝑖𝑛⁡𝑥−4 〖𝑠𝑖𝑛〗^3⁡𝑥 4 〖𝑠𝑖𝑛〗^3 𝑥=3 𝑠𝑖𝑛⁡𝑥−𝑠𝑖𝑛⁡3𝑥 〖𝑠𝑖𝑛〗^3 𝑥=(3 𝑠𝑖𝑛⁡𝑥 − 𝑠𝑖𝑛⁡3𝑥)/4 ∫1▒sin^3⁡𝑥 𝑑𝑥=∫1▒(3 sin⁡𝑥 − sin⁡3𝑥)/4 𝑑𝑥 =1/4 ∫1▒(3 sin⁡𝑥−sin⁡3𝑥 ) 𝑑𝑥 =1/4 [3∫1▒sin⁡𝑥 𝑑𝑥−∫1▒sin⁡3𝑥 𝑑𝑥] ∫1▒𝒔𝒊𝒏⁡𝟑𝒙 𝒅𝒙 Let 3𝑥=𝑡 3=𝑑𝑡/𝑑𝑥 𝑑𝑥=1/3 𝑑𝑡 ∫1▒𝒔𝒊𝒏⁡𝟑𝒙 𝒅𝒙=∫1▒sin⁡𝑡 . 1/3 𝑑𝑡 =1/3 ∫1▒sin⁡𝑡 . 𝑑𝑡 =1/3 (〖−cos〗⁡𝑡+𝐶1) =−1/3 cos⁡𝑡+1/3 𝐶1 Putting value of 𝑡 =−1/3 cos⁡3𝑥+1/3 𝐶1 ∫1▒𝒔𝒊𝒏⁡𝒙 𝒅𝒙 =−cos⁡𝑥+𝐶2 Thus, ∫1▒sin^3⁡𝑥 𝑑𝑥=1/4 [3∫1▒〖sin⁡𝑥 𝑑𝑥〗−∫1▒sin⁡3𝑥 𝑑𝑥] =1/4 [3(−cos⁡𝑥+𝐶2)−(−1/3 cos⁡3𝑥+1/3 𝐶1)] =1/4 [−3 cos⁡𝑥+3 𝐶2+ 1/3 cos⁡3𝑥+1/3 𝐶1] =1/4 [−3 cos⁡𝑥+1/3 cos⁡3𝑥+(3 𝐶2−1/3 𝐶1)] =(−3)/4 cos⁡𝑥+1/12 cos⁡3𝑥+1/4 (3 𝐶2−1/3 𝐶1) =(−𝟑)/𝟒 𝒄𝒐𝒔⁡𝒙+𝟏/𝟏𝟐 𝒄𝒐𝒔⁡𝟑𝒙+𝑪 ("As" 1/4 (3 𝐶2−1/3 𝐶1)=𝐶)