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The value of the integral`int(cos^3x+cos^5x)/(sin^2x+sin^4x)dxi s``sinx-6tan^(-1)(sinx)+C``sinx-2(sinx)^(-1)+C``sinx-2(sinx)^(-1)-6tan^(-1)(sinx)+C``sinx-2(sinx)^(-1)+5tan^(-1)(sinx)+C`A. `sin x - 6 tan^(-1)(sin x) + c`B. `sin x - 2 (sin x)^(-1) + c`C. `sin x - 2 (sin x)^(-1) - 6 tan^(-1)(sin x) + c`D. `sin x -2 (sin x)^(-1) + 5 tan^(-1) (sin x) + c` |
Answer» Correct Answer - C Let `I=int(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)dx` `=int((cos^(2)+xos^(4)x)*cos x dx)/((sin^(2)x+sin^(4)x))` Put `sin x = t rArr cos x dx = dt` `therefore" "I=int([(1-t^(2))+(1-t^(2))^(2)])/(t^(2)+t^(4))dt` `rArr" "I=int(1-t^(2)+1-2t^(2)+t^(4))/(t^(2)+t^(4))dt` `rArr" "I=int(2-3t^(2)+t^(4))/(t^(2)(t^(2)+1))dt" "....(i)` Using partial fraction for `(y^(2)-3y+2)/(y(y+1))=1+(A)/(y)+(B)/(y=1)" "["where", y = t^(2)]` `rArr" "A=2, B=-6` `therefore" "(y^(2)-3y+2)/(y(y+1))=1+(2)/(y)-(6)/(y+1)` Now, Eq. (i) reduces to, `I=int(1+(2)/(t^(2))-(6)/(1+t^(2)))dt` `=t-(2)/(t)-6 tan^(-1)(t)+c` `=sin x -(2)/(sin x)-6 tan^(-1)(sin x)+c` |
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