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cos 2x / cos^2(x) * sin^2(x) (integrals) I want to know how to solve it

Guest Jan 18, 2016

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 #2
avatar+18712 
+41

cos 2x / cos^2(x) * sin^2(x) (integrals) I want to know how to solve it

 

\(\small{ \begin{array}{rcll} \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \frac{ \cos{( 2x )} \cdot [ 1-\cos^2{(x)} ] } { \cos^2{(x)} } \qquad &| \qquad \sin^2{(x)} = 1-\cos^2{(x)} \\ &=& \cos{( 2x )} \cdot \left[ \frac{1-\cos^2{(x)}} {\cos^2{(x)}} \right] \\ &=& \cos{( 2x )} \cdot \left[ \frac{1}{\cos^2{(x)}} -1 \right] \\ &=& \frac{\cos{( 2x )} }{\cos^2{(x)}} -\cos{( 2x )} \qquad &| \qquad \cos{( 2x )} = \cos^2{(x)}- \sin^2{(x)}\\ &=& \frac{ \cos^2{(x)}- \sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{\sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{ 1-\cos^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \left[ \frac{ 1 } {\cos^2{(x)} }-1 \right] -\cos{( 2x )} \\ &=& 2- \frac{ 1 } {\cos^2{(x)} } -\cos{( 2x )} \\ \end{array} }\)

 

\(\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\ \end{array}\)

 

\(\begin{array}{rcll} \text{1.} \qquad \int 2\ dx &=& 2\int dx \\ \int 2\ dx &=& 2x \end{array}\)

 

\(\begin{array}{rcll} \text{2.} \qquad \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \int \frac{ \sin^2{(x)}+\cos^2{(x)} }{\cos^2{(x)}} \ dx \\ &=& \int (\tan^2{(x)} + 1 )\ dx\\ && \boxed{~ \begin{array}{rcll} \text{we need: } y &=& \tan{(x)} \\ y &=& \frac{\sin{(x)}} {\cos{(x)}} \\ y' &=& \frac{\sin{(x)}} {\cos{(x)}} \left[ \frac{\cos{(x)}} {\sin{(x)}} - \frac{-\sin{(x)}}{\cos{(x)}} \right] \\ y' &=& \tan{(x)} \left[ \cot{(x)} + \tan{(x)} \right] \\ y' &=& 1+ \tan^2{(x)} \\ \end{array} ~}\\ &=& \int (\tan^2{(x)} + 1 )\ dx \\ \text{we substitute:} ~ u &=& \tan{(x)}\\ du &=&\left( 1+\tan^2{(x)} \right)\ dx\\ &=& \int (\tan^2{(x)} + 1 ) \frac{du}{1+\tan^2{(x)}} \\ &=& \int du\\ &=& u\\ \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \tan{(x)}\\ \end{array}\)

 

\(\begin{array}{rcll} \text{3.} \qquad \int \cos{( 2x )}\ dx \\ \text{we substitute:} ~ u &=& 2x\\ du &=&2\ dx\\ \int \cos{( 2x )}\ dx &=& \int \cos{( u )} \frac{du}{2} \\ &=& \frac12 \cdot \int \cos{( u )} \ du \\ &=& \frac12 \cdot \sin{( u )} \\ &=& \frac12 \cdot \sin{( 2x )} \qquad &| \qquad \sin{( 2x )} = 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ &=& \frac12 \cdot 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ \int \cos{( 2x )}\ dx &=& \sin{( x )}\cdot \cos{( x )} \\ \end{array}\)

 

\(\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\\\ \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& 2x - \tan{(x)}-\sin{( x )}\cdot \cos{( x )} + c \end{array}\)

 

laugh

heureka  Jan 18, 2016
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3+0 Answers

 #1
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+15

Take the integral:
integral cos(2 x) tan^2(x) dx
For the integrand cos(2 x) tan^2(x), substitute u = tan(x) and  du = sec^2(x)  dx:
  =   integral (u^2 cos(2 tan^(-1)(u)))/(u^2+1) du
Write (u^2 cos(2 tan^(-1)(u)))/(u^2+1) as u^2/(u^2+1)^2-u^4/(u^2+1)^2:
  =   integral (u^2/(u^2+1)^2-u^4/(u^2+1)^2) du
Integrate the sum term by term and factor out constants:
  =  - integral u^4/(u^2+1)^2 du+ integral u^2/(u^2+1)^2 du
For the integrand u^4/(u^2+1)^2, do long division:
  =  - integral (-2/(u^2+1)+1/(u^2+1)^2+1) du+ integral u^2/(u^2+1)^2 du
Integrate the sum term by term and factor out constants:
  =  2 integral 1/(u^2+1) du- integral 1/(u^2+1)^2 du- integral 1 du+ integral u^2/(u^2+1)^2 du
The integral of 1/(u^2+1) is tan^(-1)(u):
  =  2 tan^(-1)(u)- integral 1/(u^2+1)^2 du- integral 1 du+ integral u^2/(u^2+1)^2 du
For the integrand 1/(u^2+1)^2, substitute u = tan(s) and  du = sec^2(s)  ds. Then (u^2+1)^2  =  (tan^2(s)+1)^2  =  sec^4(s) and s = tan^(-1)(u):
  =  2 tan^(-1)(u)- integral cos^2(s) ds- integral 1 du+ integral u^2/(u^2+1)^2 du
Write cos^2(s) as 1/2 cos(2 s)+1/2:
  =  2 tan^(-1)(u)- integral (1/2 cos(2 s)+1/2) ds- integral 1 du+ integral u^2/(u^2+1)^2 du
Integrate the sum term by term and factor out constants:
  =  2 tan^(-1)(u)-1/2 integral cos(2 s) ds-1/2 integral 1 ds- integral 1 du+ integral u^2/(u^2+1)^2 du
For the integrand cos(2 s), substitute p = 2 s and  dp = 2  ds:
  =  2 tan^(-1)(u)-1/4 integral cos(p) dp-1/2 integral 1 ds- integral 1 du+ integral u^2/(u^2+1)^2 du
The integral of cos(p) is sin(p):
  =  2 tan^(-1)(u)-(sin(p))/4-1/2 integral 1 ds- integral 1 du+ integral u^2/(u^2+1)^2 du
The integral of 1 is s:
  =  -s/2+2 tan^(-1)(u)-(sin(p))/4- integral 1 du+ integral u^2/(u^2+1)^2 du
The integral of 1 is u:
  =  -s/2-u+2 tan^(-1)(u)-(sin(p))/4+ integral u^2/(u^2+1)^2 du
For the integrand u^2/(u^2+1)^2, use partial fractions:
  =  -s/2-u+2 tan^(-1)(u)-(sin(p))/4+ integral (1/(u^2+1)-1/(u^2+1)^2) du
Integrate the sum term by term and factor out constants:
  =  -s/2-u+2 tan^(-1)(u)-(sin(p))/4+ integral 1/(u^2+1) du- integral 1/(u^2+1)^2 du
The integral of 1/(u^2+1) is tan^(-1)(u):
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4- integral 1/(u^2+1)^2 du
For the integrand 1/(u^2+1)^2, substitute u = tan(w) and  du = sec^2(w)  dw. Then (u^2+1)^2  =  (tan^2(w)+1)^2  =  sec^4(w) and w = tan^(-1)(u):
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4- integral cos^2(w) dw
Write cos^2(w) as 1/2 cos(2 w)+1/2:
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4- integral (1/2 cos(2 w)+1/2) dw
Integrate the sum term by term and factor out constants:
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4-1/2 integral cos(2 w) dw-1/2 integral 1 dw
For the integrand cos(2 w), substitute v = 2 w and  dv = 2  dw:
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4-1/4 integral cos(v) dv-1/2 integral 1 dw
The integral of cos(v) is sin(v):
  =  -s/2-u+3 tan^(-1)(u)-(sin(p))/4-(sin(v))/4-1/2 integral 1 dw
The integral of 1 is w:
  =  -(sin(p))/4-s/2-u+3 tan^(-1)(u)-(sin(v))/4-w/2+constant
Substitute back for v = 2 w:
  =  -(sin(p))/4-s/2-u+3 tan^(-1)(u)-w/2-1/4 sin(2 w)+constant
Substitute back for w = tan^(-1)(u):
  =  (-1/4 (u^2+1) (sin(p)+2 s+4 u-10 tan^(-1)(u))-u/2)/(u^2+1)+constant
Substitute back for p = 2 s:
  =  (-1/2 (u^2+1) (s+sin(s) cos(s)+2 u-5 tan^(-1)(u))-u/2)/(u^2+1)+constant
Substitute back for s = tan^(-1)(u):
  =  (-1/2 (u^2+1) (2 u-4 tan^(-1)(u)+sin(tan^(-1)(u)) cos(tan^(-1)(u)))-u/2)/(u^2+1)+constant
Simplify using cos(tan^(-1)(z)) = 1/sqrt(z^2+1) and sin(tan^(-1)(z)) = z/sqrt(z^2+1):
  =  (2 (u^2+1) tan^(-1)(u)-u (u^2+2))/(u^2+1)+constant
Substitute back for u = tan(x):
  =  -1/4 sec(x) (5 sin(x)+sin(3 x)-8 cos(x) tan^(-1)(tan(x)))+constant
Which is equivalent for restricted x values to:
Answer: | =  2 x-tan(x)-sin(x) cos(x)+constant

Guest Jan 18, 2016
 #2
avatar+18712 
+41
Best Answer

cos 2x / cos^2(x) * sin^2(x) (integrals) I want to know how to solve it

 

\(\small{ \begin{array}{rcll} \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \frac{ \cos{( 2x )} \cdot [ 1-\cos^2{(x)} ] } { \cos^2{(x)} } \qquad &| \qquad \sin^2{(x)} = 1-\cos^2{(x)} \\ &=& \cos{( 2x )} \cdot \left[ \frac{1-\cos^2{(x)}} {\cos^2{(x)}} \right] \\ &=& \cos{( 2x )} \cdot \left[ \frac{1}{\cos^2{(x)}} -1 \right] \\ &=& \frac{\cos{( 2x )} }{\cos^2{(x)}} -\cos{( 2x )} \qquad &| \qquad \cos{( 2x )} = \cos^2{(x)}- \sin^2{(x)}\\ &=& \frac{ \cos^2{(x)}- \sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{\sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{ 1-\cos^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \left[ \frac{ 1 } {\cos^2{(x)} }-1 \right] -\cos{( 2x )} \\ &=& 2- \frac{ 1 } {\cos^2{(x)} } -\cos{( 2x )} \\ \end{array} }\)

 

\(\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\ \end{array}\)

 

\(\begin{array}{rcll} \text{1.} \qquad \int 2\ dx &=& 2\int dx \\ \int 2\ dx &=& 2x \end{array}\)

 

\(\begin{array}{rcll} \text{2.} \qquad \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \int \frac{ \sin^2{(x)}+\cos^2{(x)} }{\cos^2{(x)}} \ dx \\ &=& \int (\tan^2{(x)} + 1 )\ dx\\ && \boxed{~ \begin{array}{rcll} \text{we need: } y &=& \tan{(x)} \\ y &=& \frac{\sin{(x)}} {\cos{(x)}} \\ y' &=& \frac{\sin{(x)}} {\cos{(x)}} \left[ \frac{\cos{(x)}} {\sin{(x)}} - \frac{-\sin{(x)}}{\cos{(x)}} \right] \\ y' &=& \tan{(x)} \left[ \cot{(x)} + \tan{(x)} \right] \\ y' &=& 1+ \tan^2{(x)} \\ \end{array} ~}\\ &=& \int (\tan^2{(x)} + 1 )\ dx \\ \text{we substitute:} ~ u &=& \tan{(x)}\\ du &=&\left( 1+\tan^2{(x)} \right)\ dx\\ &=& \int (\tan^2{(x)} + 1 ) \frac{du}{1+\tan^2{(x)}} \\ &=& \int du\\ &=& u\\ \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \tan{(x)}\\ \end{array}\)

 

\(\begin{array}{rcll} \text{3.} \qquad \int \cos{( 2x )}\ dx \\ \text{we substitute:} ~ u &=& 2x\\ du &=&2\ dx\\ \int \cos{( 2x )}\ dx &=& \int \cos{( u )} \frac{du}{2} \\ &=& \frac12 \cdot \int \cos{( u )} \ du \\ &=& \frac12 \cdot \sin{( u )} \\ &=& \frac12 \cdot \sin{( 2x )} \qquad &| \qquad \sin{( 2x )} = 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ &=& \frac12 \cdot 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ \int \cos{( 2x )}\ dx &=& \sin{( x )}\cdot \cos{( x )} \\ \end{array}\)

 

\(\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\\\ \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& 2x - \tan{(x)}-\sin{( x )}\cdot \cos{( x )} + c \end{array}\)

 

laugh

heureka  Jan 18, 2016
 #3
avatar+18712 
+10

Sorry:

 

\(\sin{( 2x )} = 2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}= \frac12 \cdot2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}=\sin{( x )}\cdot \cos{( x )}\\\)

 

laugh

heureka  Jan 20, 2016

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