Question

∫sin2xcos3xdx
Calculate the indefinite integral

Answer 1

Step-by-step explanation:

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I forget intergral const C pls add it

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Answer 2

Answer:

$\bold\red{\frac{1}{2} ( \cos \: x - \frac{ \cos5x }{5} ) + c}$

Step-by-step explanation:

We have to integrate,

$\int \sin(2x) \cos(3x) dx$

Multiply and divide it by 2,

we get,

$= \frac{1}{2} \int2 \sin(2x) \cos(3x) dx$

Now,

we know that,

$2 \sin( \alpha ) \cos( \beta ) = \sin( \alpha + \beta ) + \sin( \alpha - \beta )$

Therefore,

According to this identity,

we get,

$= \frac{1}{2} \int (\sin(2x + 3x) + \sin(2x - 3x) )dx \\ \\ = \frac{1}{2} \int( \sin(5x) + \sin( - x) )dx$

But,

we know that,

$\sin( - \alpha ) = - \sin( \alpha )$

Therefore,

we get,

$= \frac{1}{2} \int( \sin(5x) - \sin(x) )dx$

But,

we know that,

$\int \sin(m \alpha ) = - \frac{ \cos(m \alpha ) }{m} \\ \\ and \\ \\ \int \sin( \alpha ) = - \cos( \alpha )$

Therefore,

we get,

$= \frac{1}{2} ( - \frac{ \cos(5x) }{5} - ( - \cos(x) ) ) + c\\ \\ = \bold{\frac{1}{2} ( \cos \: x - \frac{ \cos5x }{5} ) + c}$

where,

c is an arbitrary constant