check integral is correct?

**decoy808** · June 3rd 2010, 05:16 AM

could someone check ive done this right.... many thnaks

evaluate

$ \int {sin (4x)}{cos (x)} dx $

i have....

$\int\frac{1}{2} {sin (a+b)}{sin (a-b)} dx $

$\int\frac{1}{2} {sin (5x)}{sin (3x)} dx $

$\int\frac{1}{2}\frac {(-cos (5x))}{5}+cos (3x)+c dx $

**mr fantastic** · June 3rd 2010, 05:27 AM

Originally Posted by decoy808

could someone check ive done this right.... many thnaks

evaluate

$ \int {sin (4x)}{cos (x)} dx $

i have....

$\int\frac{1}{2} {sin (a+b)}{sin (a-b)} dx $

$\int\frac{1}{2} {sin (5x)}{sin (3x)} dx $

$\int\frac{1}{2}\frac {(-cos (5x))}{5}+cos (3x)+c dx $

Check it here: integrate Sin[4x] Cos[x] - Wolfram|Alpha

Click on Show steps.

**Archie Meade** · June 3rd 2010, 05:29 AM

Originally Posted by decoy808

could someone check ive done this right.... many thnaks

evaluate

$ \int {sin (4x)}{cos (x)} dx $

i have....

$\int\frac{1}{2}\left( {sin (a+b)}\color{red}+\color{black}{sin (a-b)}\right) dx $

$\int\frac{1}{2}\left( {sin (5x)}\color{red}+\color{black}{sin (3x)}\right) dx $

$\color{red}\int\color{black}\frac{1}{2}\frac {(-cos (5x))}{5}\color{red}+cos (3x)\color{black}+c \color{red}dx $

On your last line, you've integrated the first Sine correctly but not the second one, while not removing the integration and dx symbols

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