((sin^6 x−cos^6 x)/(sin^2 xcos^2 x)).intregrate


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Question Number 49816 by Aditya789 last updated on 11/Dec/18

$\frac{\sin^{6} x - \cos^{6} x}{\sin^{2} {xcos}^{2} x} . intregrate$
	Answered by AdqhK ÐQeQqQ last updated on 11/Dec/18
	$∫(((sin^2 x)^3 −(cos^2 x)^3 )/(sin^2 x cos^2 x))dx ∫(((sin^2 x−cos^2 x)(sin^4 x+cos^4 x+sin^2 xcos^2 x))/(sin^2 x cos^2 x))dx ∫((−cos2x(1−2sin^2 xcos^2 x+sin^2 xcos^2 x))/(sin^2 xcos^2 x))dx {sin^4 x+cos^4 x=1−2sin^2 xcos^2 x} ∫(((2cos^2 x−1)(1−sin^2 xcos^2 x).4)/(4sin^2 xcos^2 x))dx ∫(((2cos^2 x−1)(4−sin^2 2x))/(sin^2 2x))dx ∫{((8cos^2 x)/(4sin^2 xcos^2 x))−((2cos^2 xsin^2 2x)/(sin^2 2x))dx−(4/(sin^2 2x))+((sin^2 2x)/(sin^2 2x))}dx ∫{2cosec^2 x−2cos^2 x−4cosec^2 x+1}dx ∫(1−2cos^2 x−2cosec^2 x)dx ∫−cos2xdx+∫−2cosec^2 xdx −(1/2)sin2x+2cotx+C pls cheak.......$
	$\int \frac{{({s i n}^{2} x)}^{3} - {({c o s}^{2} x)}^{3}}{{s i n}^{2} x {c o s}^{2} x} d x$ $\int \frac{({s i n}^{2} x - {c o s}^{2} x) ({s i n}^{4} x + {c o s}^{4} x + {s i n}^{2} {x c o s}^{2} x)}{{s i n}^{2} x {c o s}^{2} x} d x$ $\int \frac{- c o s 2 x (1 - 2 {s i n}^{2} {x c o s}^{2} x + {s i n}^{2} {x c o s}^{2} x)}{{s i n}^{2} {x c o s}^{2} x} d x {{s i n}^{4} x + {c o s}^{4} x = 1 - 2 {s i n}^{2} {x c o s}^{2} x}$ $\int \frac{(2 {c o s}^{2} x - 1) (1 - {s i n}^{2} {x c o s}^{2} x) . 4}{4 {s i n}^{2} {x c o s}^{2} x} d x$ $\int \frac{(2 {c o s}^{2} x - 1) (4 - {s i n}^{2} 2 x)}{{s i n}^{2} 2 x} d x$ $\int {\frac{8 {c o s}^{2} x}{4 {s i n}^{2} {x c o s}^{2} x} - \frac{2 {c o s}^{2} {x s i n}^{2} 2 x}{{s i n}^{2} 2 x} d x - \frac{4}{{s i n}^{2} 2 x} + \frac{{s i n}^{2} 2 x}{{s i n}^{2} 2 x}} d x$ $\int {2 {c o s e c}^{2} x - 2 {c o s}^{2} x - 4 {c o s e c}^{2} x + 1} d x$ $\int (1 - 2 {c o s}^{2} x - 2 {c o s e c}^{2} x) d x$ $\int - c o s 2 x d x + \int - 2 {c o s e c}^{2} x d x$ $- \frac{1}{2} s i n 2 x + 2 c o t x + C$ $p l s c h e a k . . . . . . .$
	Commented by arvinddayama01@gmail.com. last updated on 11/Dec/18

	$n o s i r t h i s i s w r o n g b e c a u s e$ $i t h i n k m i s t a k e s i n 4^{t h} l a s t l i n e$ $\int (2 {c o s e c}^{2} x + 1 - 2 {c o s}^{2} x - 4 {c o s e c}^{2} 2 x) d x$ $- 2 c o t x - 1 / 2 s i n 2 x + 2 c o t 2 x + C$
	Commented by AdqhK ÐQeQqQ last updated on 11/Dec/18

	$s o r r y$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
	$∫((cos^6 x(tan^6 x−1))/(tan^2 x×cos^4 x))dx ∫((tan^6 x−1)/(tan^2 x))×(1/((1+tan^2 x)))dx ∫((tan^6 x−1)/(tan^2 x))×((1+tan^2 x)/((1+tan^2 x)^2 ))dx ∫((tan^6 x−1)/(tan^2 x))×sec^2 x×(1/((1+tan^2 x)^2 ))dx k=tanx dk=sec^2 xdx ∫((k^6 −1)/k^2 )×(1/((1+k^2 )^2 ))dk ∫(((k^2 −1)(k^4 +k^2 +1))/(k^2 (k^4 +2k^2 +1)))dk ∫((1−(1/k^2 ))/)×((k^2 (k^2 +1+(1/k^2 )))/(k^2 (k^2 +2+(1/k^2 ))))dk ∫(1−(1/k^2 ))×(({(k+(1/k))^2 −2+1})/((k+(1/k))^2 ))dk p=k+(1/k) dp=(1−(1/k^2 ))dk ∫((p^2 −1)/p^2 )dp ∫dp−∫p^(−2) dp =p−((p^(−2+1) /(−2+1)))+c =p+(1/p)+c =(k+(1/k))+(1/((k+(1/k))))+c =(tanx+(1/(tanx)))+(1/((tanx+(1/(tanx)))))+c pls check....$
	$\int \frac{{c o s}^{6} x ({t a n}^{6} x - 1)}{{t a n}^{2} x \times {c o s}^{4} x} d x$ $\int \frac{{t a n}^{6} x - 1}{{t a n}^{2} x} \times \frac{1}{(1 + {t a n}^{2} x)} d x$ $\int \frac{{t a n}^{6} x - 1}{{t a n}^{2} x} \times \frac{1 + {t a n}^{2} x}{{(1 + {t a n}^{2} x)}^{2}} d x$ $\int \frac{{t a n}^{6} x - 1}{{t a n}^{2} x} \times {s e c}^{2} x \times \frac{1}{{(1 + {t a n}^{2} x)}^{2}} d x$ $k = t a n x d k = {s e c}^{2} x d x$ $\int \frac{k^{6} - 1}{k^{2}} \times \frac{1}{{(1 + k^{2})}^{2}} d k$ $\int \frac{(k^{2} - 1) (k^{4} + k^{2} + 1)}{k^{2} (k^{4} + 2 k^{2} + 1)} d k$ $\int \frac{1 - \frac{1}{k^{2}}}{} \times \frac{k^{2} (k^{2} + 1 + \frac{1}{k^{2}})}{k^{2} (k^{2} + 2 + \frac{1}{k^{2}})} d k$ $\int (1 - \frac{1}{k^{2}}) \times \frac{{{(k + \frac{1}{k})}^{2} - 2 + 1}}{{(k + \frac{1}{k})}^{2}} d k$ $p = k + \frac{1}{k} d p = (1 - \frac{1}{k^{2}}) d k$ $\int \frac{p^{2} - 1}{p^{2}} d p$ $\int d p - \int p^{- 2} d p$ $= p - (\frac{p^{- 2 + 1}}{- 2 + 1}) + c$ $= p + \frac{1}{p} + c$ $= (k + \frac{1}{k}) + \frac{1}{(k + \frac{1}{k})} + c$ $= (t a n x + \frac{1}{t a n x}) + \frac{1}{(t a n x + \frac{1}{t a n x})} + c$ $p l s c h e c k . . . .$
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