compute the definite integral of dt|cos^nt

Question

repiy me as possible as u can.

Answer 1

For a definite integration, a high and low limit for the integration need to be specified. No such information has been provided, so I'm assuming an indefinite integration.

There are various methods. A common one is the reduction method (see later). However, this method leads to having to perform multiple integrations, whittling down the exponent n by 2 each time.

METHOD 1 (for odd n)
∫cosⁿ(t)dt can be split thus: ∫(cos^n-1(t))(cos(t))dt. When n is odd it can be expressed as 2m+1, where m is an integer. n-1=2m so we have ∫cos^2m(t)cos(t)dt.
cos²(t)=1-sin²(t) so the integral becomes  ∫(1-sin²(t))^mcos(t)dt.
Let x=sin(t), dx=cos(t)dt and we get  ∫(1-x²)^mdx.
(1-x²)^m can be expanded (binomial theorem): 1-mx²+m(m-1)x⁴/2!-m(m-1)(m-2)x⁶/3!+...-x2m and then integrated:
x-mx³/3+m(m-1)x⁵/(5(2!))-m(m-1)(m-2)x⁷/(7(3!))+...+(-1)rC(m,r)x^2r+1/(2r+1)+..., where C(...,...) is the combination function C(m,r)=m!/(r!(m-r)!) for 0≤r≤m.
The coefficients can also be calculated using Pascal's Triangle:
1 1 which is C(1,0) C(1,1) for m=1
1 2 1 which is C(2,0) C(2,1) C(2,2) for m=2
1 3 3 1 which is C(3,0) C(3,1) C(3,2) C(3,3) for m=3
1 4 6 4 1 which is C(4,0) C(4,1) C(4,2) C(4,3) C(4,4) for m=4
1 5 10 10 5 1 which is C(5,0) C(5,1) C(5,2) C(5,3) C(5,4) C(5,5) for m=5
1 6 15 20 15 6 1  which is C(6,0) C(6,1) C(6,2) C(6,3) C(6,4) C(6,5) C(6,6) for m=6
...
Each row apart from the first is created from the previous row by simply adding adjacent numbers and writing the sum as the coefficient in the new row. Each row starts and ends with 1.
For example, to get from 1 3 3 1 to 1 4 6 4 1, 1+3=4, 3+3=6, 3+1=4.
The general term in the expansion is (-1)rC(m,r)x^2r+1/(2r+1) for 0≤r≤m. For example, if m=4:
(-1)0C(4,0)x/1=x; (-1)1C(4,1)x³/3=-4x³/3; (-1)2C(4,2)x⁵/5=6x⁵/5; (-1)3C(4,3)x⁷/7=-4x⁷/7; (-1)4C(4,4)x⁹/9=x⁹/9.
x is replaced by sin(t) to produce the general term (-1)rC(m,r)sin^2r+1(t)/(2r+1) for 0≤r≤m. The constant of integration C needs to be added to the result.

METHOD 2 (reduction)
Let I_n=∫cosⁿ(t)dt.

Let u=cos^n-1(t), du=-(n-1)sin(t)cos^n-2(t); dv=cos(t)dt, v=sin(t).

I_n=uv-∫vdu=sin(t)cos^n-1(t)+(n-1)∫sin²(t)cos^n-2(t)dt,

I_n=sin(t)cos^n-1(t)+(n-1)∫(1-cos²(t))cos^n-2(t)dt,

I_n=sin(t)cos^n-1(t)+(n-1)∫(cos^n-2(t)-cosⁿ(t))dt,

I_n=sin(t)cos^n-1(t)+(n-1)∫cos^n-2(t)dt-(n-1)I_n,

I_n+(n-1)I_n=sin(t)cos^n-1(t)+(n-1)∫cos^n-2(t)dt,

nI_n=sin(t)cos^n-1(t)+(n-1)∫cos^n-2(t)dt, which can be written:

nI_n=sin(t)cos^n-1(t)+(n-1)I_n-2, so I_n=(sin(t)cos^n-1(t)+(n-1)I_n-2)/n.

By repeatedly applying this eventually we arrive at I_n-2p=∫dt (n even) or ∫cos(t)dt (n odd), that is, I₀=t or I₁=sin(t). The constant of integration C can then be added.

The Reduction Method is probably best applied by starting at I₀ (even n) or I₁ (odd n) and working upward to I_n because it provides an already evaluated integral from which the next I₂, I₃, etc., can be calculated.

METHOD 3 (even n=2m, multiple angle result)

∫cosⁿ(t)dt=∫(cos²(t))^mdt=∫[(1+cos(2t))/2]^mdt=(1/2^m)∫(1+cos(2t))^mdt.

(1+cos(2t))^m=∑C(m,r)cos^r(2t) (see Method 1).

For example, if n=6, m=3:

∫cos⁶(t)dt=∫(cos²(t))³dt=∫[(1+cos(2t))/2]³dt=(1/2³)∫(1+cos(2t))³dt.

(1+cos(2t))³=1+3cos(2t)+3cos²(2t)+cos³(2t).

∫cos⁶(t)dt=∫(1+3cos(2t)+3cos²(2t)+cos³(2t))dt,

∫cos⁶(t)dt=t+3sin(2t)/2+3∫cos²(2t)dt+∫cos³(2t))dt.

∫cos²(2t)dt=½∫(1+cos(4t))dt=½(t+¼sin(4t)).

∫cos³(2t))dt=∫(1-sin²(2t))cos(2t)dt.

Let x=sin(2t), dx=2cos(2t)dt, cos(2t)dt=dx/2.

∫(1-sin²(2t))cos(2t)dt=½∫(1-x²)dx=½(x-x³/3)=½sin(2t)(1-⅓sin²(2t)).

Putting all these together:

∫cos⁶(t)dt=t+3sin(2t)/2+(3/2)(t+¼sin(4t))+½sin(2t)(1-⅓sin²(2t)).

∫cos⁶(t)dt=5t/2+2sin(2t)+⅜sin(4t)-⅙sin³(2t)+C.

Method 3 uses some of the techniques used in Method 1. As n increases and remains even the process becomes more lengthy but manageable.

These methods yield the same solution for the same problem, even though the final results may look entirely different.