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— Preceding unsigned comment added by 2402:4000:11CA:BC1F:2:2:1D15:FEA2 (talk ) 13:14, 13 February 2021 (UTC) [ reply ]
power reduction formula for
cos
3
x
{\displaystyle \cos ^{3}x}
product to sum formula for
sin
x
1
cos
x
2
{\displaystyle \sin x_{1}\cos x_{2}}
Abdul Muhsy (talk ) 14:02, 13 February 2021 (UTC) [ reply ]
d
cos
x
=
−
sin
x
d
x
.
{\displaystyle d\cos x=-\sin x\,dx.}
sin
2
x
=
2
sin
x
cos
x
{\displaystyle \sin 2x=2\sin x\cos x}
cos
2
x
=
2
cos
2
x
−
1
{\displaystyle \cos 2x=2\cos ^{2}x-1}
sin
4
x
⋅
cos
3
x
{\displaystyle \sin 4x\cdot \cos ^{3}x}
4
sin
x
cos
x
(
2
cos
2
x
−
1
)
⋅
cos
3
x
=
{\displaystyle 4\sin x\cos x(2\cos ^{2}x-1)\cdot \cos ^{3}x=}
sin
x
(
8
cos
6
x
−
4
cos
4
x
)
.
{\displaystyle \sin x(8\cos ^{6}x-4\cos ^{4}x).}
∫
sin
4
x
⋅
cos
3
x
d
x
{\displaystyle ~~~~\int \sin 4x\cdot \cos ^{3}x\,dx}
=
∫
sin
x
(
8
cos
6
x
−
4
cos
4
x
)
d
x
{\displaystyle =\int \sin x(8\cos ^{6}x-4\cos ^{4}x)\,dx}
=
∫
−
(
8
cos
6
x
−
4
cos
4
x
)
⋅
−
sin
x
d
x
,
{\displaystyle =\int -(8\cos ^{6}x-4\cos ^{4}x)\cdot -\sin x\,dx,}
=
∫
−
(
8
cos
6
x
−
4
cos
4
x
)
d
cos
x
,
{\displaystyle =\int -(8\cos ^{6}x-4\cos ^{4}x)\,d\cos x,}
=
∫
−
(
8
c
6
−
4
c
4
)
d
c
,
{\displaystyle =\int -(8c^{6}-4c^{4})\,dc,}
c
=
cos
x
,
{\displaystyle c=\cos x,}
=
−
(
8
7
c
7
−
4
5
c
5
)
{\displaystyle =-({\frac {8}{7}}c^{7}-{\frac {4}{5}}c^{5})}
=
−
(
8
7
cos
7
x
−
4
5
cos
5
x
)
.
{\displaystyle =-({\frac {8}{7}}\cos ^{7}x-{\frac {4}{5}}\cos ^{5}x).}
--Lambiam 16:33, 13 February 2021 (UTC) [ reply ] User:Abdul Muhsy 's suggestion):
∫
sin
4
x
⋅
cos
3
x
d
x
{\displaystyle \int \sin 4x\cdot \cos ^{3}x\,dx}
=
∫
sin
4
x
⋅
cos
x
⋅
cos
x
⋅
cos
x
d
x
{\displaystyle =\int \sin 4x\cdot \cos x\cdot \cos x\cdot \cos x\,dx}
=
1
2
∫
(
sin
3
x
+
sin
5
x
)
⋅
cos
x
⋅
cos
x
d
x
{\displaystyle ={\frac {1}{2}}\int (\sin 3x+\sin 5x)\cdot \cos x\cdot \cos x\,dx}
=
1
4
∫
(
sin
2
x
+
2
sin
4
x
+
sin
6
x
)
⋅
cos
x
d
x
{\displaystyle ={\frac {1}{4}}\int (\sin 2x+2\sin 4x+\sin 6x)\cdot \cos x\,dx}
=
1
8
∫
(
sin
x
+
3
sin
3
x
+
3
sin
5
x
+
sin
7
x
)
d
x
{\displaystyle ={\frac {1}{8}}\int (\sin x+3\sin 3x+3\sin 5x+\sin 7x)\,dx}
=
−
1
8
(
cos
x
+
3
cos
3
x
3
+
3
cos
5
x
5
+
cos
7
x
7
)
{\displaystyle =-{\frac {1}{8}}(\cos x+{\frac {3\cos 3x}{3}}+{\frac {3\cos 5x}{5}}+{\frac {\cos 7x}{7}})}
--RDBury (talk ) 13:03, 14 February 2021 (UTC) [ reply ]
∫
sin
(
4
x
)
cos
3
(
x
)
d
x
=
∫
e
4
i
x
−
e
−
4
i
x
2
i
(
e
i
x
+
e
−
i
x
2
)
3
d
x
=
∫
1
16
i
(
e
7
i
x
+
3
e
5
i
x
+
3
e
3
i
x
+
e
i
x
−
e
−
i
x
−
3
e
−
3
i
x
−
3
e
−
5
i
x
−
e
−
7
i
x
)
d
x
=
−
1
16
(
e
7
i
x
7
+
3
e
5
i
x
5
+
e
3
i
x
+
e
i
x
+
e
−
i
x
+
e
−
3
i
x
+
3
e
−
5
i
x
5
+
e
−
7
i
x
7
)
+
C
=
−
1
8
(
cos
(
7
x
)
7
+
3
cos
(
5
x
)
5
+
cos
(
3
x
)
+
cos
(
x
)
)
+
C
{\displaystyle {\begin{aligned}\int \sin(4x)\cos ^{3}(x)dx&=\int {\frac {e^{4ix}-e^{-4ix}}{2i}}\left({\frac {e^{ix}+e^{-ix}}{2}}\right)^{3}dx\\&=\int {\frac {1}{16i}}(e^{7ix}+3e^{5ix}+3e^{3ix}+e^{ix}-e^{-ix}-3e^{-3ix}-3e^{-5ix}-e^{-7ix})dx\\&=-{\frac {1}{16}}\left({\frac {e^{7ix}}{7}}+3{\frac {e^{5ix}}{5}}+e^{3ix}+e^{ix}+e^{-ix}+e^{-3ix}+3{\frac {e^{-5ix}}{5}}+{\frac {e^{-7ix}}{7}}\right)+C\\&=-{\frac {1}{8}}\left({\frac {\cos(7x)}{7}}+{\frac {3\cos(5x)}{5}}+\cos(3x)+\cos(x)\right)+C\end{aligned}}}
(though
−
8
7
cos
7
(
x
)
+
4
5
cos
5
(
x
)
{\displaystyle -{\frac {8}{7}}\cos ^{7}(x)+{\frac {4}{5}}\cos ^{5}(x)}
catslash (talk ) 19:18, 14 February 2021 (UTC) [ reply ] Tamfang (talk ) 02:22, 15 February 2021 (UTC) [ reply ] Lambiam 12:09, 15 February 2021 (UTC) [ reply ] could differ by a constant (they do not)
cos
(
7
x
)
=
64
cos
7
(
x
)
−
112
cos
5
(
x
)
+
56
cos
3
(
x
)
−
7
cos
(
x
)
{\displaystyle \scriptstyle {\cos(7x)=64\cos ^{7}(x)-112\cos ^{5}(x)+56\cos ^{3}(x)-7\cos(x)}}
cos
(
5
x
)
=
16
cos
5
(
x
)
−
20
cos
3
(
x
)
+
5
cos
(
x
)
{\displaystyle \scriptstyle {\cos(5x)=16\cos ^{5}(x)-20\cos ^{3}(x)+5\cos(x)}}
cos
(
3
x
)
=
4
cos
3
(
x
)
−
3
cos
(
x
)
{\displaystyle \scriptstyle {\cos(3x)=4\cos ^{3}(x)-3\cos(x)}}
cos
7
(
x
)
=
1
64
(
cos
(
7
x
)
+
7
cos
(
5
x
)
+
21
cos
(
3
x
)
+
35
cos
(
x
)
)
{\displaystyle \scriptstyle {\cos ^{7}(x)={\frac {1}{64}}(\cos(7x)+7\cos(5x)+21\cos(3x)+35\cos(x))}}
cos
5
(
x
)
=
1
16
(
cos
(
5
x
)
+
5
cos
(
3
x
)
+
10
cos
(
x
)
)
{\displaystyle \scriptstyle {\cos ^{5}(x)={\frac {1}{16}}(\cos(5x)+5\cos(3x)+10\cos(x))}}
cos
3
(
x
)
=
1
4
(
cos
(
3
x
)
+
3
cos
(
x
)
)
{\displaystyle \scriptstyle {\cos ^{3}(x)={\frac {1}{4}}(\cos(3x)+3\cos(x))}}
catslash (talk ) 13:28, 15 February 2021 (UTC) [ reply ]
x
=
π
/
2.
{\displaystyle x=\pi /2.}
Lambiam 14:54, 15 February 2021 (UTC) [ reply ] Tamfang (talk ) 01:37, 18 February 2021 (UTC) [ reply ] 
