solve-the-integral-sin-4-d-

Question Number 25965 by Chuks” last updated on 17/Dec/17

s o l v e t h e i n t e g r a l

\int \sin^{4} θ d θ

Answered by $@ty@m last updated on 17/Dec/17

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

\Rightarrow \sin^{3} x = \frac{1}{4} (3 \sin x - \sin 3 x)

∴ T h e g i v e n i n t e g r a l :

\int \sin^{4} x d x

= \int \sin^{3} x . \sin x d x

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

= \frac{3}{4} \int \sin^{2} x d x - \frac{1}{8} \int 2 \sin 3 x \sin x d x

= \frac{3}{4} \int \frac{1 - \cos 2 x}{2} d x - \frac{1}{8} \int (\cos 2 x - \cos 4 x) d x

= \frac{3}{8} \int d x - \frac{3}{8} \int \cos 2 x d x - \frac{1}{8} \int \cos 2 x d x + \frac{1}{8} \int \cos 4 x d x

= \frac{3}{8} x - \frac{3}{16} \sin 2 x - \frac{1}{16} \sin 2 x + \frac{1}{32} \sin 4 x + C

Commented by Chuks” last updated on 29/Dec/17

w o w

G r e a t w o r k . .

M o r e k n o w l e d g e s i r

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