How to evaluate the integral? I used substitution but my answer was wrong. dx /

Question

How to evaluate the integral?

I used substitution but my answer was wrong.

Explanation / Answer

Substitute 6x+1 = y^2 => 6 dx= 2y dy Integral becomes int ((12/(y^2-1)^2)dy) Substitute y = secp dy = secp tanp dp Integral becomes int (12 cos^2 p/sin^3 p) dp Use cos^2 p = 1 - sin^2 p Integral becomes int (12 (1 -sin^2 p)/sin^3 p) dp = int (12cosec^3 p - 12 cosec p) dp Now int (cosec^3 p) dp : Let u = cosec p, v = -cotp du = -cosecp cotp dp, dv = cosec^2 p dp int (cosec^3 p) dp = - cosecp cotp - int cosec p cot^2p dp = - cosecp cotp - int cosecp (cosec^2 p - 1) dp int cosec^3 p dp = = - cosecp cotp - int (cosec^3p - cosecp) dp 2 int cosec^3p dp = - cosecp cotp + ln|cosecp - cotp| + C int cosec^3p dp = -1/2 cosecp cotp + 1/2 ln|cosecp - cotp| + C And int (cosec p) dp: To find this, cosec p = 1/sinp = sinp/sin^2p = (sin p)/(1 - cos^2p). Let u = cos p and then proceed by partial fractions after the substitution. We get, int (cosec p)dp = - ln|cosecp+cotp| +C Therefore, int (12cosec^3 p - 12 cosec p) dp = 12 (-1/2 cosecp cotp + 1/2 ln|cosecp - cotp|) - 12 (- ln|cosecp+cotp|) +C = -6 cosecpcotp + 6 ln|cosecp - cotp| +12 ln|cosecp+cotp| + C Now, substitute back y=secp and then 6x+1 = y^2 Hope this helps. Revert in case of any need for clarifications. Thanks.

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