find: ∫_0 ^Π (cos^6 θ −cos^4 θ) dθ plase help me in cinding this And also explain if possible


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Question Number 52164 by pooja24 last updated on 04/Jan/19

$f i n d :$ $\underset{0}{\int^{Π}} ({c o s}^{6} θ - \cos^{4} θ) d θ$ $p l a s e h e l p m e i n c i n d i n g t h i s A n d a l s o$ $e x p l a i n i f p o s s i b l e$
Commented by prakash jain last updated on 04/Jan/19

$\int_{0}^{π / 2} \sin^{m} x \cos^{n} x d x$ $= \frac{[(n - 1) (n - 3) . . 1 or 2]] (m - 1) (m - 3) . . 1 o r 2]}{(m + n) (m + n - 2) . . . 1 or 2} K$ $where K = \frac{π}{2} if both m, n are even$ $K = 1 otherwise$ $I = \int_{0}^{π / 2} \cos^{6} x d x - \int_{0}^{π / 2} \cos^{4} x d x$ $+ \int_{π / 2}^{π} \cos^{6} x d x - \int_{π / 2}^{π} \cos^{4} x d x$ $\int_{π / 2}^{π} \cos^{6} x d x = \int_{0}^{π / 2} \sin^{6} x d x = \int_{0}^{π / 2} \cos^{6} x d x$ $\int_{π / 2}^{π} \cos^{4} x d x = \int_{0}^{π / 2} \sin^{4} x d x = \int_{0}^{π / 2} \cos^{4} x d x$ $I = 2 [\int_{0}^{π / 2} \cos^{6} x d x - \int_{0}^{π / 2} \cos^{4} x d x]$ $= 2 [\frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} - \frac{3 \cdot 1}{4 \cdot 2}] \frac{π}{2} ∵ m = 6 even$ $= 2 [\frac{5}{16} - \frac{3}{8}] \frac{π}{2}$ $= - \frac{π}{16}$
Commented by maxmathsup by imad last updated on 04/Jan/19

$l e t I = \int_{0}^{π} ({c o s}^{6} x - {c o s}^{4} x) d x \Rightarrow I = \int_{0}^{π} ({c o s}^{4} x (1 - {s i n}^{2} x - {c o s}^{4} x) d x$ $= - \int_{0}^{π} {s i n}^{2} x {c o s}^{4} x d x = - \int_{0}^{π} \frac{1 - c o s (2 x)}{2} {(\frac{1 + c o s (2 x)}{2})}^{2} d x$ $= - \frac{1}{8} \int_{0}^{π} (1 - {c o s}^{2} (2 x)) (1 + c o s (2 x)) d x$ $= - \frac{1}{8} \int_{0}^{π} (1 - \frac{1 + c o s (4 x)}{2}) (1 + c o s (2 x)) d x$ $= - \frac{1}{8} \int_{0}^{π} (\frac{1 - c o s (4 x)}{2}) (1 + c o s (2 x)) d x$ $= - \frac{1}{16} \int_{0}^{π} (1 + c o s (2 x) - c o s (4 x) - c o s (2 x) c o s (4 x)) d x$ $= - \frac{π}{16} - \frac{1}{16} \int_{0}^{π} c o s (2 x) d x + \frac{1}{16} \int_{0}^{π} c o s (4 x) d x + \frac{1}{16} \int_{0}^{π} c o s (2 x) c o s (4 x) d x$ $= - \frac{π}{16} - \frac{1}{32} {[s i n (2 x)]}_{0}^{π} + \frac{1}{64} {[s i n (4 x)]}_{0}^{π} + \frac{1}{32} \int_{0}^{π} (c o s (6 x) + c o s (2 x)) d x$ $= - \frac{π}{16} + 0 + 0 + 0 = - \frac{π}{16} .$
Commented by maxmathsup by imad last updated on 04/Jan/19
$error of typo at first line I =∫_0 ^π {cos^4 x(1−sin^2 x)−cos^4 x}dx...$
$e r r o r o f t y p o a t f i r s t l i n e I = \int_{0}^{π} {{c o s}^{4} x (1 - {s i n}^{2} x) - {c o s}^{4} x} d x . . .$
Commented by pooja24 last updated on 05/Jan/19

$t h a n k y o u$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19

	$- {c o s}^{4} θ (1 - {c o s}^{2} θ)$ $= - {c o s}^{4} θ {s i n}^{2} θ$ $= - {(\frac{1 + c o s 2 θ}{2})}^{2} (\frac{1 - c o s 2 θ}{2})$ $= - \frac{1}{8} (1 + c o s 2 θ) (1 - {c o s}^{2} 2 θ)$ $= - \frac{1}{8} (1 + c o s 2 θ) (1 - \frac{1 + c o s 4 θ}{2})$ $= \frac{- 1}{16} (1 + c o s 2 θ) (2 - 1 - c o s 4 θ)$ $= \frac{- 1}{16} (1 + c o s 2 θ) (1 - c o s 4 θ)$ $= \frac{- 1}{16} (1 - c o s 4 θ + c o s 2 θ - c o s 2 θ c o s 4 θ)$ $= \frac{- 1}{16} (1 - c o s 4 θ + c o s 2 θ - \frac{c o s 6 θ + c o s 2 θ}{2})$ $= \frac{- 1}{32} (2 - 2 c o s 4 θ + 2 c o s 2 θ - c o s 6 θ - c o s 2 θ)$ $= \frac{1}{32} (c o s 6 θ + 2 c o s 4 θ - c o s 2 θ - 2)$ $n o w$ $\frac{1}{32} \int_{0}^{π} c o s 6 θ + 2 c o s 4 θ - c o s 2 θ - \frac{1}{16} \int_{0}^{π} d θ$ $\frac{1}{32} ∣ \frac{s i n 6 θ}{6} + \frac{2 s i n 4 θ}{4} - \frac{s i n 2 θ}{2} ∣_{0}^{π} - \frac{1}{16} \times π$ $\underset{}{} \underset{i n t r e g a l v a l u e z e r \underset{}{o}}{⇓} - \frac{1}{16} \times π$ $a n s w e r i s = \frac{- π}{16} [s i n n π = 0]$ $p l s c h e k . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19

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