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Question Number 88388 by M±th+et£s last updated on 10/Apr/20

Answered by mind is power last updated on 10/Apr/20

49sec^2 (x)+28tan(x)+9sin^2 (x)−6sin(x)−44  =49(1+tg^2 (x))+28tg(x)+(3sin(x)−1)^2 −45  =4+49tg^2 (x)+28tg(x)+(3sin(x)−1)^2   =(7tg(x)+2)^2 +(3sin(x)−1)^2   21sin(x)+6cos(x)−21tg(x)sec(x)+7sec^2 (x)  =21sin(x)+6cos(x)−21sin(x)(1+tg^2 (x))+7(1+tg^2 (x))  =6cos(x)+7+7tg^2 (x)(1−3sin(x))  =∫((7tg^2 (1−3sin(x))+6cos(x)+7)/((7tg(x)+2)^2 +(3sin(x)−1)^2 ))dx  =∫(((7tg^2 (1−3sin(x))+6cos(x)+7)/((1−3sin(x))^2 ))/(1+(((7tg(x)+2)/(−3sin(x)+1)))^2 ))dx    7tg^2 (x)(1−3sin(x))+6cos(x)+7  =7tg^2 (x)(1−3sin(x))+7(1−3sin(x))+3cos(x)(7tg(x)+2)  =7(1+tg^2 (x))(1−3sin(x))+3cos(x)(7tg(x)+2)  ⇔∫((7((d(7tg(x)+2)(1−3sin(x))−d(1−3sin(x))(7tg(x)+2))/((1−3sin(x))^2 )))/(1+(((7tg(x)+2)/(1−3sin(x))))^2 ))dx  =∫((d(((7tg(x)+2)/(1−3sin(x)))))/(1+(((7tg(x)+2)/(1−3sin(x))))^2 ))dx  =tan^(−1) (((7tg(x)+2)/(1−3sin(x))))+c

49sec2(x)+28tan(x)+9sin2(x)6sin(x)44=49(1+tg2(x))+28tg(x)+(3sin(x)1)245=4+49tg2(x)+28tg(x)+(3sin(x)1)2=(7tg(x)+2)2+(3sin(x)1)221sin(x)+6cos(x)21tg(x)sec(x)+7sec2(x)=21sin(x)+6cos(x)21sin(x)(1+tg2(x))+7(1+tg2(x))=6cos(x)+7+7tg2(x)(13sin(x))=7tg2(13sin(x))+6cos(x)+7(7tg(x)+2)2+(3sin(x)1)2dx=7tg2(13sin(x))+6cos(x)+7(13sin(x))21+(7tg(x)+23sin(x)+1)2dx7tg2(x)(13sin(x))+6cos(x)+7=7tg2(x)(13sin(x))+7(13sin(x))+3cos(x)(7tg(x)+2)=7(1+tg2(x))(13sin(x))+3cos(x)(7tg(x)+2)7d(7tg(x)+2)(13sin(x))d(13sin(x))(7tg(x)+2)(13sin(x))21+(7tg(x)+213sin(x))2dx=d(7tg(x)+213sin(x))1+(7tg(x)+213sin(x))2dx=tan1(7tg(x)+213sin(x))+c

Commented by john santu last updated on 10/Apr/20

amazing sir

amazingsir

Commented by M±th+et£s last updated on 10/Apr/20

great solution

greatsolution

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