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Evaluate \int \sin ^{2}x\cos ^{4}xdx..

 

  • Option 1)

    \frac{\sin 6x}{192}-\frac{\sin 4x}{64}-\frac{1}{64}\sin 2x+\frac{x}{16}+c

  • Option 2)

    -\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x+\frac{x}{16}+c

  • Option 3)

    -\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x-\frac{x}{16}+c

  • Option 4)

    \frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x-\frac{x}{16}+c

 

Answers (1)

As we learnt

Different cases of type of indefinite integration -

(sin^{m}x)\left ( cos^{n} x\right ) \therefore \int \left ( sin^{m}xcos^{n}x \right )dx

Case     m      n

I          Odd    Even

II         Even   Odd

III        Odd     Odd

IV        Even   Even

- wherein

Substitution

I    Put t=cosx

II    Put t=sinx

III   Put t=sinx or t=cosx

iV  If (m+n)<0  put  t=tan x  or, If (m+n)>0  use DeMoivre's Theorem.

 

 We have \,\int{{{\sin }^{2}}x{{\cos }^{4}}}\,x\,dx \,=\,\,\int{\left( \frac{1\,-\,\cos 2x}{2} \right)\,\,{{\left( \frac{\cos 2x+1}{2} \right)}^{2}}\,dx}

                \begin{array}{l} = \,\frac{1}{8}\int {\left( {1\, - \,\cos 2x} \right)\,({{\cos }^2}} 2x\, + \,2\,\cos 2x\, + \,1)\,dx\,\,\,\\ \, = \,\,\frac{1}{8}\int {\left( {\frac{{\cos 4x\, + \,1}}{2}} \right)\,dx\, + \,\int {\frac{{2\cos \,2xdx}}{8}\, + \,\int {\frac{{dx}}{8}} } } \,\, - \,\frac{1}{8}\int {{{\cos }^3}\,2xdx\,} \end{array}

                   - \,\frac{1}{4}\,\int {{{\cos }^2}2xdx\, - \,\frac{1}{8}\,\int {\cos 2xdx} }

I\; = \;\frac{1}{{16}}\;\frac{{\sin 4x}}{4}\; + \;\frac{x}{{16}} + \;\frac{{\sin 2x}}{8}\; + \,\frac{x}{8}\; - \frac{1}{8}\;\int {\frac{{\cos \;6x\, + 3\;\cos 2x}}{4}\;dx\;}

- \frac{1}{8}\;\left[ {(\cos \;4x + 1)\;dx\; - \;\frac{1}{{16}}\;\sin \;2x} \right]

= \;\frac{1}{{64}}\;\sin 4x\; + \;\frac{1}{{16}}\sin \;2x\; + \;\frac{{3x}}{{16}}\; - \frac{1}{{32}}\;\left[ {\frac{{\sin \;6x}}{6}\; + \;\frac{{3\;\sin \;2x}}{2}} \right]\; - \frac{1}{{32}}\sin \;4x\; - \;\frac{x}{8}

= \; - \frac{{\sin \;6x}}{{192}}\; - \;\frac{{\sin \;4x}}{{64}}\; + \frac{1}{{64\;}}\;\sin \;2x\; + \frac{x}{{16}}\; + \;c.

 


Option 1)

\frac{\sin 6x}{192}-\frac{\sin 4x}{64}-\frac{1}{64}\sin 2x+\frac{x}{16}+c

Option 2)

-\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x+\frac{x}{16}+c

Option 3)

-\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x-\frac{x}{16}+c

Option 4)

\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x-\frac{x}{16}+c

Posted by

Vakul

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