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February 13

Integration

How to integrate sin 4 x.cos ^ { 3 } x d x — Preceding unsigned comment added by 2402:4000:11CA:BC1F:2:2:1D15:FEA2 (talk) 13:14, 13 February 2021 (UTC)

\cos^{3} x

\sin x_{1} \cos x_{2}

. -Abdul Muhsy (talk) 14:02, 13 February 2021 (UTC)

Observe that

d \cos x = - \sin x d x .

This can be exploited in this specific case for a change of variable, giving a slightly less laborious calculation. Use the double-angle formulas

\sin 2 x = 2 \sin x \cos x

and

\cos 2 x = 2 \cos^{2} x - 1

repeatedly to rewrite

\sin 4 x \cdot \cos^{3} x

as

4 \sin x \cos x (2 \cos^{2} x - 1) \cdot \cos^{3} x =

\sin x (8 \cos^{6} x - 4 \cos^{4} x) .

Then

\int \sin 4 x \cdot \cos^{3} x d x

= \int \sin x (8 \cos^{6} x - 4 \cos^{4} x) d x

= \int - (8 \cos^{6} x - 4 \cos^{4} x) \cdot - \sin x d x,

= \int - (8 \cos^{6} x - 4 \cos^{4} x) d \cos x,

= \int - (8 c^{6} - 4 c^{4}) d c,

where

c = \cos x,

= - (\frac{8}{7} c^{7} - \frac{4}{5} c^{5})

= - (\frac{8}{7} \cos^{7} x - \frac{4}{5} \cos^{5} x) .

--Lambiam 16:33, 13 February 2021 (UTC)

Just for fun, another method (expanding on User:Abdul Muhsy's suggestion):

\int \sin 4 x \cdot \cos^{3} x d x

= \int \sin 4 x \cdot \cos x \cdot \cos x \cdot \cos x d x

= \frac{1}{2} \int (\sin 3 x + \sin 5 x) \cdot \cos x \cdot \cos x d x

= \frac{1}{4} \int (\sin 2 x + 2 \sin 4 x + \sin 6 x) \cdot \cos x d x

= \frac{1}{8} \int (\sin x + 3 \sin 3 x + 3 \sin 5 x + \sin 7 x) d x

= - \frac{1}{8} (\cos x + \frac{3 \cos 3 x}{3} + \frac{3 \cos 5 x}{5} + \frac{\cos 7 x}{7})

--RDBury (talk) 13:03, 14 February 2021 (UTC)

Brute force:

\begin{matrix} \int \sin (4 x) \cos^{3} (x) d x & = \int \frac{e^{4 i x} - e^{- 4 i x}}{2 i} {(\frac{e^{i x} + e^{- i x}}{2})}^{3} d x \\ = \int \frac{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 \\ = - \frac{1}{16} (\frac{e^{7 i x}}{7} + 3 \frac{e^{5 i x}}{5} + e^{3 i x} + e^{i x} + e^{- i x} + e^{- 3 i x} + 3 \frac{e^{- 5 i x}}{5} + \frac{e^{- 7 i x}}{7}) + C \\ = - \frac{1}{8} (\frac{\cos (7 x)}{7} + \frac{3 \cos (5 x)}{5} + \cos (3 x) + \cos (x)) + C \end{matrix}

(though

- \frac{8}{7} \cos^{7} (x) + \frac{4}{5} \cos^{5} (x)

is surely a neater way of expressing this). catslash (talk) 19:18, 14 February 2021 (UTC)

\cos (7 x) = 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)

\cos (3 x) = 4 \cos^{3} (x) - 3 \cos (x)

\cos^{7} (x) = \frac{1}{64} (\cos (7 x) + 7 \cos (5 x) + 21 \cos (3 x) + 35 \cos (x))

\cos^{5} (x) = \frac{1}{16} (\cos (5 x) + 5 \cos (3 x) + 10 \cos (x))

\cos^{3} (x) = \frac{1}{4} (\cos (3 x) + 3 \cos (x))