Ex 7.3, 10 - Chapter 7 Class 12 Integrals

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Ex 7.3, 10 Integrate the function 𝑠𝑖𝑛4 𝑥 ∫1▒sin^4⁡𝑥 𝑑𝑥 =∫1▒(sin^2⁡𝑥 )^2 𝑑𝑥 =∫1▒((1 − cos⁡2𝑥)/2)^2 𝑑𝑥 =1/4 ∫1▒(1−cos⁡2𝑥 )^2 𝑑𝑥 We know that 𝑐𝑜𝑠⁡2𝜃=1−2 〖𝑠𝑖𝑛〗^2⁡𝜃 2 〖𝑠𝑖𝑛〗^2⁡𝜃=1−𝑐𝑜𝑠⁡2𝜃 〖𝑠𝑖𝑛〗^2⁡𝜃=(1 − 𝑐𝑜𝑠⁡2𝜃)/2 Replace 𝜃 by 𝑥 〖𝑠𝑖𝑛〗^2⁡𝑥=(1 − 𝑐𝑜𝑠⁡2𝑥)/2 =1/4 ∫1▒(1^2+(cos⁡2𝑥 )^2−2(1)(cos⁡2𝑥 )) 𝑑𝑥 =1/4 ∫1▒(1+cos^2⁡2𝑥−2 cos⁡2𝑥 ) 𝑑𝑥 =1/4 ∫1▒(1+(1 + cos⁡4𝑥)/2−2 cos⁡2𝑥 ) 𝑑𝑥 =1/4 ∫1▒1 𝑑𝑥+1/8 ∫1▒(1+cos⁡4𝑥 ) 𝑑𝑥−2/4 ∫1▒cos⁡2𝑥 𝑑𝑥 =1/4 ∫1▒1 𝑑𝑥+1/8 ∫1▒1 𝑑𝑥+1/8 ∫1▒cos⁡4𝑥 𝑑𝑥−1/2 ∫1▒cos⁡2𝑥 𝑑𝑥 We know that cos⁡2𝜃=2 cos^2⁡〖𝜃−1〗 cos⁡〖2𝜃+1〗=2 cos^2⁡𝜃 Replace 𝜃 by 2𝑥 cos⁡〖4𝑥+1〗=2 cos^2⁡2𝑥 cos⁡〖4𝑥 + 1〗/2=cos^2⁡2𝑥 ∫1▒cos⁡(𝑎𝑥+𝑏) 𝑑𝑥=𝑠𝑖𝑛⁡(𝑎𝑥 + 𝑏)/𝑎 =𝑥/4 + 𝑥/8 + 1/8 sin⁡4𝑥/4 − 1/2 sin⁡2𝑥/2 +𝐶 =(2𝑥 + 𝑥)/8 + 1/32 sin⁡4𝑥− 1/4 sin⁡2𝑥+𝐶 =𝟑𝒙/𝟖− 𝟏/𝟒 𝒔𝒊𝒏⁡𝟐𝒙++ 𝟏/𝟑𝟐 𝒔𝒊𝒏⁡𝟒𝒙+𝑪