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B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 21, 2010 #3 Evaluate 7∫sin[ln∣8x∣]dx.\displaystyle Evaluate \ 7\int sin[ln|8x|]dx.Evaluate 7∫sin[ln∣8x∣]dx. OK, let k = ln∣8x∣, dk = dxx, and x = ek8.\displaystyle OK, \ let \ k \ = \ ln|8x|, \ dk \ = \ \frac{dx}{x}, \ and \ x \ = \ \frac{e^k}{8}.OK, let k = ln∣8x∣, dk = xdx, and x = 8ek. Therefore, 7∫sin[ln∣8x∣]dx = 78∫sin(k)ekdk, remember this.\displaystyle Therefore, \ 7\int sin[ln|8x|]dx \ = \ \frac{7}{8}\int sin(k)e^kdk, \ remember \ this.Therefore, 7∫sin[ln∣8x∣]dx = 87∫sin(k)ekdk, remember this. Now 78∫sin(k)ekdk, let u = sin(k), du = cos(k)dk\displaystyle Now \ \frac{7}{8}\int sin(k)e^kdk, \ let \ u \ = \ sin(k), \ du \ = \ cos(k)dkNow 87∫sin(k)ekdk, let u = sin(k), du = cos(k)dk and dv = ekdk, ⟹ v = ek.\displaystyle and \ dv \ = \ e^kdk, \ \implies \ v \ = \ e^k.and dv = ekdk, ⟹ v = ek. Hence, we have 78∫sin(k)ekdk = 78eksin(k)−78∫ekcos(k)dk.\displaystyle Hence, \ we \ have \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}\int e^kcos(k)dk.Hence, we have 87∫sin(k)ekdk = 87eksin(k)−87∫ekcos(k)dk. Now, ∫ekcos(k)dk = ekcos(k)+∫eksin(k)dk.\displaystyle Now, \ \int e^kcos(k)dk \ = \ e^kcos(k)+\int e^ksin(k)dk.Now, ∫ekcos(k)dk = ekcos(k)+∫eksin(k)dk. Ergo, we now have: 78∫sin(k)ekdk = 78eksin(k)−78ekcos(k)−78∫eksin(k)dk.\displaystyle Ergo, \ we \ now \ have: \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)-\frac{7}{8}\int e^ksin(k)dk.Ergo, we now have: 87∫sin(k)ekdk = 87eksin(k)−87ekcos(k)−87∫eksin(k)dk. Hence, 74∫eksin(k)dk = 78eksin(k)−78ekcos(k)+C or ∫eksin(k)dk = eksin(k)2−ekcos(k)2+C.\displaystyle Hence, \ \frac{7}{4}\int e^ksin(k)dk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)+C \ or \ \int e^ksin(k)dk \ = \ \frac{e^ksin(k)}{2}-\frac{e^kcos(k)}{2}+C.Hence, 47∫eksin(k)dk = 87eksin(k)−87ekcos(k)+C or ∫eksin(k)dk = 2eksin(k)−2ekcos(k)+C. \(\displaystyle \int e^ksin(k)dk \ = \ 8\int sin[ln|8x|}dx \ = \ 4xsin[ln|8x|]-4xcos[ln|8x|] +C.\) Therefore 7∫sin[ln∣8x∣]dx = 7x2[sin(ln∣8x∣)−cos(ln∣8x∣)]+C, QED.\displaystyle Therefore \ 7\int sin[ln|8x|]dx \ = \ \frac{7x}{2}[sin(ln|8x|)-cos(ln|8x|)]+C, \ QED.Therefore 7∫sin[ln∣8x∣]dx = 27x[sin(ln∣8x∣)−cos(ln∣8x∣)]+C, QED. Ryan, where do you get these "Icky" problems?\displaystyle Ryan, \ where \ do \ you \ get \ these \ "Icky" \ problems?Ryan, where do you get these "Icky" problems? If these are the problems your Professor gives you, then he or she must indeed\displaystyle If \ these \ are \ the \ problems \ your \ Professor \ gives \ you, \ then \ he \ or \ she \ must \ indeedIf these are the problems your Professor gives you, then he or she must indeed be a stern taskmaster.\displaystyle be \ a \ stern \ taskmaster.be a stern taskmaster.
Evaluate 7∫sin[ln∣8x∣]dx.\displaystyle Evaluate \ 7\int sin[ln|8x|]dx.Evaluate 7∫sin[ln∣8x∣]dx. OK, let k = ln∣8x∣, dk = dxx, and x = ek8.\displaystyle OK, \ let \ k \ = \ ln|8x|, \ dk \ = \ \frac{dx}{x}, \ and \ x \ = \ \frac{e^k}{8}.OK, let k = ln∣8x∣, dk = xdx, and x = 8ek. Therefore, 7∫sin[ln∣8x∣]dx = 78∫sin(k)ekdk, remember this.\displaystyle Therefore, \ 7\int sin[ln|8x|]dx \ = \ \frac{7}{8}\int sin(k)e^kdk, \ remember \ this.Therefore, 7∫sin[ln∣8x∣]dx = 87∫sin(k)ekdk, remember this. Now 78∫sin(k)ekdk, let u = sin(k), du = cos(k)dk\displaystyle Now \ \frac{7}{8}\int sin(k)e^kdk, \ let \ u \ = \ sin(k), \ du \ = \ cos(k)dkNow 87∫sin(k)ekdk, let u = sin(k), du = cos(k)dk and dv = ekdk, ⟹ v = ek.\displaystyle and \ dv \ = \ e^kdk, \ \implies \ v \ = \ e^k.and dv = ekdk, ⟹ v = ek. Hence, we have 78∫sin(k)ekdk = 78eksin(k)−78∫ekcos(k)dk.\displaystyle Hence, \ we \ have \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}\int e^kcos(k)dk.Hence, we have 87∫sin(k)ekdk = 87eksin(k)−87∫ekcos(k)dk. Now, ∫ekcos(k)dk = ekcos(k)+∫eksin(k)dk.\displaystyle Now, \ \int e^kcos(k)dk \ = \ e^kcos(k)+\int e^ksin(k)dk.Now, ∫ekcos(k)dk = ekcos(k)+∫eksin(k)dk. Ergo, we now have: 78∫sin(k)ekdk = 78eksin(k)−78ekcos(k)−78∫eksin(k)dk.\displaystyle Ergo, \ we \ now \ have: \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)-\frac{7}{8}\int e^ksin(k)dk.Ergo, we now have: 87∫sin(k)ekdk = 87eksin(k)−87ekcos(k)−87∫eksin(k)dk. Hence, 74∫eksin(k)dk = 78eksin(k)−78ekcos(k)+C or ∫eksin(k)dk = eksin(k)2−ekcos(k)2+C.\displaystyle Hence, \ \frac{7}{4}\int e^ksin(k)dk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)+C \ or \ \int e^ksin(k)dk \ = \ \frac{e^ksin(k)}{2}-\frac{e^kcos(k)}{2}+C.Hence, 47∫eksin(k)dk = 87eksin(k)−87ekcos(k)+C or ∫eksin(k)dk = 2eksin(k)−2ekcos(k)+C. \(\displaystyle \int e^ksin(k)dk \ = \ 8\int sin[ln|8x|}dx \ = \ 4xsin[ln|8x|]-4xcos[ln|8x|] +C.\) Therefore 7∫sin[ln∣8x∣]dx = 7x2[sin(ln∣8x∣)−cos(ln∣8x∣)]+C, QED.\displaystyle Therefore \ 7\int sin[ln|8x|]dx \ = \ \frac{7x}{2}[sin(ln|8x|)-cos(ln|8x|)]+C, \ QED.Therefore 7∫sin[ln∣8x∣]dx = 27x[sin(ln∣8x∣)−cos(ln∣8x∣)]+C, QED. Ryan, where do you get these "Icky" problems?\displaystyle Ryan, \ where \ do \ you \ get \ these \ "Icky" \ problems?Ryan, where do you get these "Icky" problems? If these are the problems your Professor gives you, then he or she must indeed\displaystyle If \ these \ are \ the \ problems \ your \ Professor \ gives \ you, \ then \ he \ or \ she \ must \ indeedIf these are the problems your Professor gives you, then he or she must indeed be a stern taskmaster.\displaystyle be \ a \ stern \ taskmaster.be a stern taskmaster.
