caseclosed 10 Posted December 16, 2005 Share Posted December 16, 2005 integrate sin(4x)^2 I know sin(x)^2=1-cos(x)^2 cos(x)^2=sin(x)^2+cos(2x) ok I am lost, problem is sin(4x) so but the identities are in sin(x) Link to post Share on other sites

TD 10 Posted December 16, 2005 Share Posted December 16, 2005 The problem isn't the 4x, but the square! Since you can easily integrate sin(x), it's no problem to integrate sin(ax) with a an arbitrary constant either! To get rid of the square, use another trig identity: [math]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}[/math] Now you lost the square, went to a cosine and doubled the angle - but that shouldn't be a problem Link to post Share on other sites

caseclosed 10 Posted December 22, 2005 Author Share Posted December 22, 2005 umm, sin(x)^2= (1-cos(x))/2 but what does sin(4x)^2=? I am very confused on what to do with stuff inside the trig function since I can't pull it out in front Link to post Share on other sites

PhDP 76 Posted December 22, 2005 Share Posted December 22, 2005 [math]\sin^2(x) = \frac{1-\cos(2x)}{2}[/math] You just have to replace the "x" in the identity with your "4x"; [math]\sin^2(4x) = \frac{1-\cos(2(4x))}{2}=\frac{1-\cos(8x)}{2}[/math] So your integral is now; [math]\int \sin^2(4x)dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos(8x)dx[/math] Link to post Share on other sites

Dave 251 Posted December 22, 2005 Share Posted December 22, 2005 Phil: I think you may have meant [imath]\sin^2 4x[/imath] instead of [imath]\sin^2 x[/imath] in your last integral. As for evaluating something like [math]\int\cos(ax)\, dx[/math], note the following: [math]\frac{d}{dx} \sin(ax) = a\cos(ax)[/math] Now, integrate both sides to get: [math]\sin(ax) = a\int\cos(ax)\, dx[/math] You should be able to see where to go from here Link to post Share on other sites

caseclosed 10 Posted December 22, 2005 Author Share Posted December 22, 2005 You just have to replace the "x" in the identity with your "4x"; thanks, that is the part I was really confused on for a really really long time. Link to post Share on other sites

PhDP 76 Posted December 22, 2005 Share Posted December 22, 2005 Phil: I think you may have meant \sin^2 4x instead of \sin^2 x in your last integral. Exactly, but I can't correct it now, we're doomed ! Link to post Share on other sites

Dave 251 Posted December 23, 2005 Share Posted December 23, 2005 Corrected it for you Link to post Share on other sites

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