Sin(3x)cos(5x)Solution To integrate sin3xcos.pdf
Sin(3x)cos(5x)Solution To integrate sin3xcos.pdf
Sec theta + csc theta/tan theta + cot theta = sin theta + cos theta sin3 theta + cos3 theta/sin theta + cos theta = 1 - sin theta cos theta 1 + sin theta + cos theta/1 + sin theta - cos theta = 1 + cos theta/sin theta Solution 1) (sec a + csc a) /( tan a+cot a) = (1/cos a +1/sin a) /(sin a/cos a+ cos a/sin a) = (sin a+cos a / sina cos a) /(sin^2 a + cos ^2 a / sin a cos a) but sin ^2 a+cos ^2 a=1 so = sin a+cos a =RHS hence proved 2)using the formula a^3+b^3=(a+b)(a^2+b^2-ab) we solve LHS =(sin^ 3 a+cos^3 a)/(sin a+cos a) =(sin a+cos a)(sin^2 a+cos^2 a-sin a cosa ) /(sin a+cos a) =(1-sin a*cos a) =RHS hence proved 3) (1+sin a+cos a) /(1+sin a- cos a) multiply with conjugate up and down =( 1+sin a +cos a)/(1+sin a-cos a) * (1+sin a +cos a)/(1+sin a+cos a) = (1+cos ^2 a+2cos a+sin ^2 a) /(1+sin^2 a+2sin a-cos ^2 a) = (1+1 +2 cos a)/(cos ^2 a +2 sin a -cos ^2 a) =(2+2 cos a)/(2 sin a) =(1+cos a)/sin a.
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Sin(3x)cos(5x)Solution To integrate sin3xcos.pdf
- 1. Sin(3x)cos(5x) Solution To integrate sin3x*cos5x . We know that Int {u(x)v(x)} = u(x)Int v(x)dx - Int {u'(x) Intv(x) dx}dx. Let I = Int (sin3x *cos5x) dx = sin3x (sin5x)/5 - (1/5) Int ((sin3x )' sin5x)dx I = (1/5) sin3x sin5x - (1/5) Int (3cos3x*sin5x) dx I = (1/5)sin3xsin5x - (3/5){cos3x* (-cos5x)/5 - Int (-3sin3x)(-cos5x/5)}dx I = (1/5) sin3xsin5x +(3/25)(cos3x+cos5x) + (9/25) I (1- 9/25)I = (1/5)sin3xsin5x +(3/25)(cos3x+cos5x) Divide both sides by (1-9/25) = 16/25: I = (5/16)(sin3xsin5x) +(3/16)(cos3x+cos5x). Therefore Int (sin3x*cos3x)dx = (5/16)(sin3xsin5x) +(3/16)(cos3x+cos5x).