1. Class 12
2. Important Question for exams Class 12
3. Chapter 7 Class 12 Integrals

Transcript

Ex 7.3, 6 ๐ ๐๐ ๐ฅ sinโก2๐ฅ sinโก3๐ฅ โซ1โsinโกใ๐ฅ sinโกใ2๐ฅ sinโก3๐ฅ ใ ใ ๐๐ฅ =โซ1โใ(sinโก๐ฅ sinโก2๐ฅ ) sinโก3๐ฅ ใ ๐๐ฅ We know that 2 sinโก๐ด sinโก๐ต=โcosโก(๐ด+๐ต)+cosโก(๐ดโ๐ต) sinโก๐ด sinโก๐ต=1/2 [โcosโก(๐ด+๐ต)+cosโก(๐ดโ๐ต) ] sinโก๐ด sinโก๐ต=1/2 [cosโก(๐ดโ๐ต)โcosโก(๐ด+๐ต) ] Replace A by ๐ฅ & B by 2๐ฅ sinโก๐ฅ sinโก3๐ฅ=1/2 [cosโก(๐ฅโ2๐ฅ)โcosโก(๐ฅ+2๐ฅ) ] sinโก๐ฅ sinโก3๐ฅ =1/2 [cosโก(โ๐ฅ)โcosโก(3๐ฅ) ] sinโก๐ฅ sinโก3๐ฅ =1/2 [cosโก(๐ฅ)โcosโก3๐ฅ ] Thus, โซ1โsinโกใ๐ฅ sinโกใ2๐ฅ sinโก3๐ฅ ใ ใ ๐๐ฅ =โซ1โใ1/2 [(cosโก๐ฅโcosโก3๐ฅ )] ใ . sinโก3๐ฅ.๐๐ฅ =1/2 โซ1โ(cosโก๐ฅโcosโก3๐ฅ ) sinโก3๐ฅ ๐๐ฅ =1/2 [โซ1โ(cosโก๐ฅ. sinโก3๐ฅโcosโก3๐ฅ. sinโก3๐ฅ ) ]๐๐ฅ =1/2 [โซ1โใcosโก๐ฅ. sinโก3๐ฅ ใ ๐๐ฅโโซ1โใcosโก3๐ฅ. sinโก3๐ฅ ใ ๐๐ฅ] โซ1โใ๐๐๐โก๐. ๐๐๐โก๐๐ ใ ๐๐ We know that 2 ๐ ๐๐โก๐ด ๐๐๐ โก๐ต =๐ ๐๐โก(๐ด+๐ต)+๐ ๐๐โก(๐ดโ๐ต) ๐ ๐๐โก๐ด ๐๐๐ โก๐ต =1/2 [๐ ๐๐โก(๐ด+๐ต)+๐ ๐๐โก(๐ดโ๐ต) ] Replace A by 3๐ฅ & B by ๐ฅ sinโก3๐ฅ cosโก๐ฅ = 1/2 [๐ ๐๐โก(๐ฅ+3๐ฅ)+sinโก(3๐ฅโ๐ฅ) ] = 1/2 [๐ ๐๐โก4๐ฅ+sinโก2๐ฅ ] โซ1โใ๐๐๐โก๐๐. ๐๐๐โก๐๐ ใ ๐๐ We know that 2 ๐ ๐๐โก๐ด ๐๐๐ โก๐ต =๐ ๐๐โก(๐ด+๐ต)+๐ ๐๐โก(๐ดโ๐ต) ๐ ๐๐โก๐ด ๐๐๐ โก๐ต =1/2 [๐ ๐๐โก(๐ด+๐ต)+๐ ๐๐โก(๐ดโ๐ต) ] Replace A by 3๐ฅ & B by 3๐ฅ sinโก3๐ฅ cosโก3๐ฅ = 1/2 [๐ ๐๐โก(3๐ฅ+3๐ฅ)+sinโก(3๐ฅโ3๐ฅ) ] = 1/2 [๐ ๐๐โก6๐ฅ+sinโก0 ]=1/2 [๐ ๐๐โก6๐ฅ ] Hence โซ1โใsinโก3๐ฅ.cosโก๐ฅ ใ ๐๐ฅ =1/2 โซ1โ[๐ ๐๐โก4๐ฅ+sinโก2๐ฅ ] ๐๐ฅ Hence โซ1โใcosโก3๐ฅ.sinโก3๐ฅ ใ ๐๐ฅ =1/2 โซ1โsinโก6๐ฅ ๐๐ฅ Thus, our equation becomes โซ1โsinโกใ๐ฅ sinโกใ2๐ฅ sinโก3๐ฅ ใ ใ ๐๐ฅ =1/2 [โซ1โใsinโก3๐ฅ cosโก3๐ฅ ใ ๐๐ฅโโซ1โใsinโก3๐ฅ cosโก3๐ฅ ใ ๐๐ฅ] =1/2 [1/2 โซ1โ(sinโก4๐ฅ+sinโก2๐ฅ ) ๐๐ฅโ1/2 โซ1โ(sinโก6๐ฅ ) ๐๐ฅ] =1/4 [โซ1โ(sinโก4๐ฅ+sinโก2๐ฅ ) ๐๐ฅโโซ1โ(sinโก6๐ฅ ) ๐๐ฅ] =1/4 [โซ1โsinโก4๐ฅ ๐๐ฅ+โซ1โsinโก2๐ฅ ๐๐ฅโโซ1โsinโก6๐ฅ ๐๐ฅ] โซ1โsinโก(๐๐ฅ+๐) ๐๐ฅ=โ๐๐๐ โก(๐๐ฅ + ๐)/๐ +๐ถ =1/4 [(โcosโก4๐ฅ)/4 +(ใโcosใโก2๐ฅ/2) โ((โcosโก6๐ฅ)/6)]+๐ถ =1/4 [(โcosโก4๐ฅ)/4 โcosโก2๐ฅ/2+cosโก6๐ฅ/6]+๐ถ =๐/๐ [๐๐๐โก๐๐/๐ โ๐๐๐โก๐๐/๐ โ ๐๐๐โก๐๐/๐ ]+๐ช

Chapter 7 Class 12 Integrals

Class 12
Important Question for exams Class 12