without beta function ∫cos^3 t sin^2 t dt


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Question Number 64320 by aliesam last updated on 16/Jul/19

$w i t h o u t b e t a f u n c t i o n$ $\int {c o s}^{3} t {s i n}^{2} t d t$
Commented by mathmax by abdo last updated on 16/Jul/19

$I = \int s i n t (s i n t {c o s}^{3} t) d t b y p a r t s u = s i n t a n d v^{^{'}} = s i n t {c o s}^{3} t$ $\Rightarrow I = - \frac{1}{4} s i n t {c o s}^{4} t + \frac{1}{4} \int {c o s}^{5} t d t + c$ $\int {c o s}^{5} t d t = \int c o s t {({c o s}^{2} t)}^{2} d t = \int c o s t {(\frac{1 + c o s (2 t)}{2})}^{2} d t$ $= \int \frac{1}{4} c o s t (1 + 2 c o s t + {c o s}^{2} (2 t)) d t$ $= \frac{1}{4} \int c o s t d t + \frac{1}{2} \int {c o s}^{2} t d t + \frac{1}{4} \int c o s t {c o s}^{2} (2 t) d t$ $= \frac{1}{4} s i n t + \frac{1}{4} \int (1 + c o s (2 t)) d t + \frac{1}{8} \int c o s t (1 + c o s (4 t)) d t$ $= \frac{1}{4} s i n t + \frac{t}{4} + \frac{1}{8} s i n (2 t) + \frac{1}{8} s i n t + \frac{1}{8} \int c o s t c o s (4 t) d t$ $= \frac{3}{8} s i n t + \frac{t}{4} + \frac{1}{8} s i n (2 t) + \frac{1}{8} \int c o s t c o s (4 t) d t$ $= \frac{3}{8} s i n t + \frac{t}{4} + \frac{1}{8} s i n (2 t) + \frac{1}{16} \int (c o s (5 t) + c o s (3 t)) d t$ $= \frac{3}{8} s i n t + \frac{t}{4} + \frac{1}{8} s i n (2 t) + \frac{1}{80} s i n (5 t) + \frac{1}{48} s i n (3 t) \Rightarrow$ $I = - \frac{1}{4} s i n t {c o s}^{4} t + \frac{3}{32} s i n t + \frac{t}{16} + \frac{1}{24} s i n (2 t) + \frac{1}{320} s i n (5 t) + \frac{1}{192} s i n (3 t) + c$
	Answered by Tanmay chaudhury last updated on 16/Jul/19

	$\int (1 - {s i n}^{2} t) {s i n}^{2} t c o s t d t$ $\int (a^{2} - a^{4}) d a a = s i n t d a = c o s t d t$ $\frac{a^{3}}{3} - \frac{a^{5}}{5} + c$ $\frac{{s i n}^{3} t}{3} - \frac{{s i n}^{5} t}{5} + c$
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