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Evaluate \int \sin ^{4}xdx

  • Option 1)

    \frac{1}{8}+\frac{\sin 4x}{32}+ \frac{\sin 2x}{2}+C

  • Option 2)

    \frac{1}{8}-\frac{\sin 4x}{32}-\frac{\sin 2x}{2}+C

  • Option 3)

    \frac{1}{8}-\frac{\sin 4x}{32}+ \frac{\sin 2x}{2}+C

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we have learned

Integration of trigonometric function of power m -

\int sin^{m}xdx    and 

 \int cos^{m}xdx

- wherein

for m=2,

sin^{2}x=\frac{1-cos 2x}{2}

for m=3,

sin3x=3sinx-4sin^{3}x

 

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

=\int \left ( 1/4+ \frac{\cos 2x}{4}- \cos 2x \right )dx

=\int \left ( 1/4+ \frac{1+ \cos 4x}{8}- \cos 2x \right )dx

1/8 - \frac{\sin 4x}{32 }- \frac{\sin 2x}{2}+ C

 

 

 

 


Option 1)

\frac{1}{8}+\frac{\sin 4x}{32}+ \frac{\sin 2x}{2}+C

This is incorrect

Option 2)

\frac{1}{8}-\frac{\sin 4x}{32}-\frac{\sin 2x}{2}+C

This is correct

Option 3)

\frac{1}{8}-\frac{\sin 4x}{32}+ \frac{\sin 2x}{2}+C

This is incorrect

Option 4)

none of these

This is incorrect

Posted by

gaurav

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