Ex 7.3
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Ex 7.3, 5
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Ex 7.3, 23 (MCQ)
Ex 7.3, 24 (MCQ) Important
Last updated at April 16, 2024 by Teachoo
Ex 7.3, 1 Find the integral of sin2 (2𝑥 + 5) ∫1▒〖𝒔𝒊𝒏𝟐 (𝟐𝒙 + 𝟓) 〗 𝒅𝒙 =∫1▒(1 − 〖𝑐𝑜𝑠 2〗(2𝑥 + 5))/2 𝑑𝑥 =1/2 ∫1▒〖1−cos(4𝑥+10) 〗 𝑑𝑥 =1/2 [∫1▒1 𝑑𝑥−∫1▒cos(4𝑥+10) 𝑑𝑥] We know that 𝐜𝐨𝐬 𝟐𝜽=𝟏−𝟐 〖𝒔𝒊𝒏〗^𝟐𝜽 2 sin^2 𝜃=1−cos2𝜃 sin^2 𝜃=1/2 [1−cos2𝜃 ] Replace 𝜃 by (𝟐𝐱+𝟓) sin^2 (2𝑥+5)=(1 − cos2(2𝑥 + 5))/2 As ∫1▒cos(𝑎𝑥+𝑏) 𝑑𝑥=sin(𝑎𝑥 + 𝑏)/𝑎+𝐶 =1/2 [𝑥− sin(4𝑥 + 10)/4 +𝐶] =𝒙/𝟐 − 𝟏/𝟖 𝒔𝒊𝒏(𝟒𝒙+𝟏𝟎)+𝑪