2014年12月19日 · See explanation. This will be a long answer. So what you want to find is: int cos^6(x)dx There's a rule of thumb that you can remember: whenever you need to integrate an even power of the cosine function, you need to use the identity: cos^2(x) = (1+cos(2x))/2 First we split up the cosines: int cos^2(x)*cos^2(x)*cos^2(x) dx Now we can replace every cos^2(x) with the identity above: int (1+cos ...

2018年5月23日 · LHS=sin^6x+cos^6x =(sin^2x)^3+(cos^2x)^3 Using formula a^3+b^3=(a+b)^3-3ab(a+b) =(sin^2x+cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x) =1^3+3sin^2xcos^2x*1 =1-3sin^2xcos^2x=RHS

2016年1月30日 · int cos^6 x dx = 5/16 x + 1/4 sin 2x + 3/16 sin 4x -1/48 sin^3 2x int cos^6x dx We can rewrite this function as; int (cos^2x)^3dx Now we can rewrite cos^2 x using the half angle formula. int(1/2(1+cos(2x)))^3 dx If we expand the expression we can get rid of the exponent. 1/8 int (1 + 3cos(2x) + 3cos^2 (2x) + cos^3 (2x) )dx We can split up this expression using the sum rule. 1/8 int dx + 3/8 ...

2017年1月3日 · Using identity #a^3-b^3=(a-b)(a^2+b^2-ab)#. #sin^6x−cos^6x# = #(sin^2x)^3-(cos^2x)^3# = #(sin^2x-cos^2x)((sin^2x)^2+(cos^2x)^2-sin^2xcos^2x)#

2018年7月29日 · int cos^6x dx = ( 8cos^5x sinx +10 cos^3x sinx +15cosxsinx+15x)/48+C Write the integrand as: cos^6x = cos^5x * cosx So: int cos^6x dx = int cos^5x * cosx dx Integrate by parts: int cos^6x dx = int cos^5x d/dx (sinx) dx int cos^6x dx = cos^5x sinx - int sinx d/dx (cos^5x) dx int cos^6x dx = cos^5x sinx + 5 int sin^2x cos^4x dx Use now the identity: sin^2x = 1-cos^2x int cos^6x dx = cos^5x sinx ...

2016年6月20日 · Recall that #cos^2theta=(1+cos2theta)/2.# #:. cos^2(6x)=(1+cos12x)/2.# #:. intcos^2(6x)dx=int(1+cos12x)/2dx=(1/2){x+(sin12x)/12}+C.#

Frequency and period are related inversely. A period #P# is related to the frequency #f# # P = 1/f#. Something that repeats once per second has a period of 1 s.

2018年6月6日 · pi/6; pi/4; pi/3; (3pi)/4; (+ (kpi)/2) cos 6x + cos 4x + cos 2x = 0 (1) Use trig identity: cos (a + b) = cos ((a + b)/2)cos ((a - b)/2) We have: cos 6x + cos 2x = 2cos 4xcos 2x Equation (1) becomes: 2cos 4x.cos 2x + cos 2x = 0 cos 2x(2cos 4x + 1) = 0 Either factor should be zero. a. cos 2x = 0 Unit circle gives 2 solutions for 2x 1. 2x = pi/2 + 2kpi, - …

2018年1月18日 · How do you find the derivative of #sin(cos(6x))#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 2 Answers

Look at the coefficients of the function and the function variable respectively. The "period" is the distance or time it takes for the function to complete one full cycle - 360^o. The "amplitude" is the height of the curve, or the distance from the baseline (midpoint) to the highest or lowest point. The amplitude is determined by the coefficient on the function - 5 in this case. The sign ...