Integral Sin 4 3X Cos 3X Dx. PDF file{4{3(a)We use usubstitution Let u = x2 9 so du = 2xdx and 1 2 du = xdx The integral becomes Z x p x2 9 dx = 1 2 Z 1 p u du = 1 2 Z u 1 2 du 1 2 u1 2 1 2 +C = p x2 9+C (b)This is an arcsecant type integral with k = 3 so.
intsin^3(3x)cos(3x)dx=frac{sin^4(3x)}{12}+c Take the given equation intsin^3(3x)cos3xdx Take the function sin(3x)=t Now differentiate with respect to t on both sides of this \frac{d}{dt}(sin3x)=1\implies3cos3x*\frac{dx}{dt}=1\impliescos3xdx=dt/3 Substituting the given above values for the main equation we get intt^3/3dt This looks easy.
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Integral of cos^3(x)/sin^4(x) How to integrate it step by step using the substitution method! Youtube https//wwwyoutubecom/integralsforyou?sub_confirm.
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Plus 1/2 plus 1/2 cosine of four x dx Let’s see I could take this 1/2 and add it to this one and that’s going to get me 3/2 so add those together I’m going to get 3/2 So let me rewrite this as I’m in the home stretch really this is going to be equal to 1/4 times the integral of 3/2 minus two cosine of two x.