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 …

2018年1月18日 · I don't think that #cos(x^6)# has an antiderivative expressible with elementary functions, so I assume you want to find #int\ cos^6(x)\ dx#. First, recall that …

2017年2月18日 · #int cos^6xdx = sinxcos^5x + 5 int cos^4x - 5int cos^6x dx# As the integral is on both sides we can solve for it: #6 int cos^6xdx = sinxcos^5x + 5 int cos^4xdx #

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 …

"The fundamental trigonometric identities" are the basic identities: •The reciprocal identities •The pythagorean identities

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions.

2018年3月9日 · We want to expand . #sin^6(x)# One way is to use these identities repeatedly . #sin^2(x)=1/2(1-cos(2x))# #cos^2(x)=1/2(1+cos(2x))#

2018年5月30日 · -6sin(x)cos(x)^5 first you take the derivative as normal which is 6*cos(x)^5 then by the chain rule you take the derivative of the inner function which is cosin in this case and …

2018年3月9日 · We'll use rarra^3+b^3=(a+b)(a^2-ab+b^2) rarra^2+b^2=(a-b)^2+2ab rarrsin^2x+cos^2x=1 rarr2cos^2x=1+cos2x and rarr2sin^2x=1-cos2x LHS=cos^6(x)+sin^6(x) …

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 …