How to transform the result of this integral into powers of $\sin\,x$?

Question

I want to calculate the integral $I = \displaystyle\int{\cos ^5 x\, \mathrm dx}$. Using the variable $t = \sin x$, I get $$I = \sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x + C.$$ With Mathematica, when I input

 Integrate[Cos[x]^5, x]

I get

(5 Sin[x])/8 + 5/48 Sin[3 x] + 1/80 Sin[5 x]

How to format the result in the form? $$I = \sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x.$$

Similar questions with integral $I =\displaystyle \int{\dfrac{\mathrm{d}x}{1 + \sin x + \cos x}}$ by setting $t = \tan\dfrac{x}{2}$, we have $I = \ln \left|1 + \tan\dfrac{x}{2}\right|$. I tried

(Integrate[1/(1 + Sin[x] - Cos[x]), x] // TrigExpand) /. {Sin[x] -> 
     2*Tan[x/2]/(1 + Tan[x/2]^2), 
    Cos[x] -> (1 - Tan[x/2]^2)/(1 + Tan[x/2]^2)} // 
  Simplify // TraditionalForm

and get the result log(sin(x/2)+cos(x/2))-log(cos(x/2)). How to format the result in the form?
$$I = \ln \left|1 + \tan\dfrac{x}{2}\right|$$.

Answer 1

If you expand your expression in terms of elementary trig functions

Integrate[Cos[x]^5, x] // TrigExpand

(5 Sin[x])/8 + 5/16 Cos[x]^2 Sin[x] + 1/16 Cos[x]^4 Sin[x] - (
 5 Sin[x]^3)/48 - 1/8 Cos[x]^2 Sin[x]^3 + Sin[x]^5/80

you will see only even powers of Cos - so it's straightforward to replace it like this:

(Integrate[Cos[x]^5, x] // TrigExpand) /. 
   Cos[x] -> Sqrt[1 - Sin[x]^2] // Simplify // TraditionalForm

Answer 2

I'll answer with a more general module I've done for converting trig expressions. An overkill here, but anyway:

trigSet[exp_, inTerm_] := 
 Module[{trigSyms, rels, set, setRep, setRep1, toLow, oneInTermsOf, allInTermsOf, fq, 
          ruleAll, convert},

  trigSyms = {Sin, Cos, Tan, Cot, Sec, Csc};
  rels     = {csc sin == 1, cos^2 + sin^2 == 1, 1 == cos sec, tan == sin/cos, cot tan == 1};
  set      = ToExpression /@ ToLowerCase /@ SymbolName /@ trigSyms;
  setRep   = Thread[set -> (ToExpression /@ (StringJoin[#, "[x_]"] & /@ ToString /@ set))];
  setRep1  = Thread[set -> (ToExpression /@ (StringJoin[#, "[x]"] & /@ ToString /@ set))];
  toLow    = Thread[trigSyms -> set];

  oneInTermsOf[one_, of_] :=  Solve[rels, {one}, Complement[set, {one, of}]];
  allInTermsOf[of_] :=  Flatten[oneInTermsOf[#, of] & /@ Complement[set, {of}]];
  fq[x_, y_] := FreeQ[x, Alternatives @@ Complement[set, {y}]];
  ruleAll[of_] := Rule @@@ Transpose[{#[[1]] /. setRep, #[[2]] /. setRep1} &@
                                                     Transpose@(List @@@ allInTermsOf[of])];
  convert[expr_, inTerms_] := FullSimplify@ Union@Select[
      Flatten@NestWhile[# /. (List /@ ruleAll[inTerms]) &, {TrigExpand[expr] /. 
           toLow }, ! Or @@ (fq[#, inTerms] & /@ Flatten@#) &], fq[#, inTerms] &];
  HoldForm[ Evaluate@convert[exp, inTerm]] /. (Reverse /@ toLow)
  ]

  trigSet[(5 Sin[x])/8 + 5/48 Sin[3 x] + 1/80 Sin[5 x], sin]
  (*
  -> {Sin[x]-(2 Sin[x]^3)/3+Sin[x]^5/5}
  *)

Test it with more difficult expressions:

trigSet[Cos[3 x] - Tan[2 x] + Cot[3 x]^2, sin]

Edit

Depending on your expression, because of the signs in the radicals, you could need more than one "converted" one to cover the whole domain. For example:

s = Cos[x] Sin[x];
s0 = trigSet[s, sin]
s1 = FullSimplify[Reduce[# == s, x, Reals] & /@ ReleaseHold[s0] /. _Equal -> False]
(*
{-Sin[x] Sqrt[1-Sin[x]^2],Sin[x] Sqrt[1-Sin[x]^2]}

{C[1] \[Element] Integers && 
      ((Pi + x > 2 Pi  C[1] && Pi + 2 x <= 4 Pi C[1]) || Pi/2 <= x - 2 Pi C[1] < Pi),
 C[1] \[Element] Integers &&
       -(Pi/2) <= x - 2 \[Pi] C[1] <= Pi/2}

*)

Still working in automating this last process, but you can plot it like:

f[x_] := Piecewise[{{-Sin[x] Sqrt[1 - Sin[x]^2], 
                      Resolve[Exists[C[1], Element[C[1], Integers], 
                              ((Pi + x > 2 Pi  C[1] && Pi + 2 x <= 4 Pi C[1]) || 
                                Pi/2 <= x - 2 Pi C[1] < Pi)]]}, 
                    {Sin[x] Sqrt[1 - Sin[x]^2], 
                       Resolve[Exists[C[1], Element[C[1], Integers],
                             -(Pi/2) <= x - 2 \[Pi] C[1] <= Pi/2]]}}];
Plot[{f[x], Sin[x] Cos[x]}, {x, 0, 2 Pi}, PlotStyle -> {{Thick, Dashed, Red}, Blue, Green}]

Answer 3

It's somewhat convenient to use Chebyshev polynomials for such purposes. Here, I use the Chebyshev polynomial of the second kind, due to the convenient identity

$$U_n(\cos\,x)=\frac{\sin((n+1)x)}{\sin\,x}$$

Here goes:

Integrate[Cos[x]^5, x] /.
       Sin[n_Integer x] :> Sin[x] ChebyshevU[n - 1, Sqrt[1 - Sin[x]^2]] // Expand
   Sin[x] - (2 Sin[x]^3)/3 + Sin[x]^5/5

For completeness, the relation $T_n(\cos\,x)=\cos\,nx$ can be similarly exploited. I use that in this answer.

Answer 4

One can use Sin[x]^2 + Cos[x]^2 == 1 and the third argument in Solve to eliminate an unwanted dependent variable (i.e. Cos[x]) :

Expand @ Solve[{L == TrigExpand @ Integrate[Cos[x]^5, x], Sin[x]^2 + Cos[x]^2 == 1},
                L, {Cos[x]}]

That is useful in more general cases, but with the problem at hand we can observe, that we needn't eliminate Cos[x], e.g. :

Expand [Solve[{L= TrigExpand @ Integrate[Cos[x]^5, x], Sin[x]^2 + Cos[x]^2 == 1}, L]][[1, 1, 2]]

Sin[x] - (2 Sin[x]^3)/3 + Sin[x]^5/5

A bit more reliable variation of Vitaliy's approach :

TrigExpand[ Integrate[Cos[x]^5, x]] /. Solve[Sin[x]^2 + Cos[x]^2 == 1, Cos[x]] // 
Simplify // Union

Another approach : let's substitute Cos[x]^n for an even integer n by (1 - Sin[x]^2)^(n/2), e.g.

Simplify[ TrigExpand[ Integrate[Cos[x]^5, x]] /. Cos[x]^n_?EvenQ -> (1 - Sin[x]^2)^(n/2)]

Edit

To proceed with another problem of the OP remembering that Integrate[1/(1 + Sin[x] - Cos[x]), x] and Log @ Abs[1 + Tan[x/2]] are not quite equivalent, we have to assume some restrictions of the variable x. Let x be real in the range where Cos[x/2] + Sin[x/2] > 0 and Cos[x/2] > 0, e.g. Pi > x > -Pi/2, now we have :

Simplify[ Integrate[ 1/(1 + Sin[x] + Cos[x]), x], Pi > x > -Pi/2]

Log[1 + Tan[x/2]]

% // TraditionalForm