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Find the integral \int x \sin (3x^{2}+2)dx

  • Option 1)

    \frac{\cos (3x^{2}+2)}{6} +C

  • Option 2)

    -\frac{\cos (3x^{2}+2)}{6} +C

  • Option 3)

    \frac{\cos (3x^{2}+2)}{3} +C

  • Option 4)

    -\frac{\cos (3x^{2}+2)}{3} +C

 

Answers (1)

As we have learned

Type of integration by substitution -

Integral of the functions containing functions of trigonometric functions 

\because \int sin f(x)\left f'(x) \right dx

\therefore \int sint dt = -cos t+c

ex:  \therefore \int sin\left ( ax+b \right )dx=\frac{-cos(ax+b)}{a}+c

- wherein

Let f(x)=t

\therefore f'(x)dx=dt

 

Put 3x^{2}+2= t

6xdx= dt \Rightarrow xdx =dt/6

Thus I = \int \sin t dt/6= \frac{-\cos (3x^{2}+2)}{6}+ C 

 

 

 

 


Option 1)

\frac{\cos (3x^{2}+2)}{6} +C

This is incorrect

Option 2)

-\frac{\cos (3x^{2}+2)}{6} +C

This is correct

Option 3)

\frac{\cos (3x^{2}+2)}{3} +C

This is incorrect

Option 4)

-\frac{\cos (3x^{2}+2)}{3} +C

This is incorrect

Posted by

Vakul

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