sin-3-7x-dx-

Question Number 12464 by tawa last updated on 23/Apr/17

\int \sin^{3} (7 x) dx

Answered by ajfour last updated on 23/Apr/17

\sin 3 x = 3 \sin x - 4 \sin^{3} x

I = \int \sin^{3} (7 x) d x

= \frac{1}{4} \int [3 \sin (7 x) - \sin 3 (7 x)] d x

= \frac{1}{4} (- \frac{3}{7} \cos (7 x) + \frac{\cos (21 x)}{21}) + C

= - \frac{3 \cos (7 x)}{28} + \frac{\cos (21 x)}{84} + C

Commented by tawa last updated on 23/Apr/17

God bless you sir .

Answered by sma3l2996 last updated on 23/Apr/17

= \int (1 - {c o s}^{2} (7 x)) s i n (7 x) d x

= \int s i n (7 x) d x - \int s i n (7 x) {c o s}^{2} (7 x) d x

= \frac{- 1}{7} c o s (7 x) + \frac{1}{7} \int {c o s}^{'} (7 x) {c o s}^{2} (7 x) d x + c

= \frac{1}{3 \times 7} {c o s}^{3} (7 x) - \frac{1}{7} c o s (7 x)

= \frac{c o s (7 x)}{7} (\frac{1}{3} {c o s}^{2} (7 x) - 1)

Commented by tawa last updated on 23/Apr/17

God bless you sir

Answered by mrW1 last updated on 23/Apr/17

\int \sin^{3} (7 x) dx

= \frac{1}{7} \int \sin^{3} (7 x) d (7 x)

= \frac{1}{7} \int \sin^{3} t d t t = 7 x

= - \frac{1}{7} \int \sin^{2} t d (\cos t)

= - \frac{1}{7} \int (1 - \cos^{2} t) d (\cos t)

= - \frac{1}{7} \int (1 - u^{2}) d u u = \cos t = \cos (7 x)

= - \frac{1}{7} (\int d u - \int u^{2} d u)

= - \frac{1}{7} (u - \frac{u^{3}}{3}) + C

= \frac{u}{21} (u^{2} - 3) + C

= \frac{\cos (7 x)}{21} [\cos^{2} (7 x) - 3] + C

Commented by tawa last updated on 23/Apr/17

God bless you sir

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