Question

please help me with this problem
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Answer 1

Given to evaluate:-

• $\int \sin2x \cos3x \: dx$

Solution:-

$\int \sin2x \cos3x \: dx$

Using,

$\sin(t) \cos(s) = \frac{1}{2} ( \sin(t + s) + \sin(t - s) )$

We get,

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

Now, using the identity,

$\sin( - x) = - \sin(x)$

We get,

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

Using the property of integrals,

$\int a \times f(x) \: dx = a \int f(x) \: dx,a \in \R$

We get,

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

Now, using the property of integrals,

$\int f(x) \pm g(x) = \int f(x) \: dx \pm \int g(x) \: dx$

We get,

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

Now, evaluate the indefinite integral,

$\frac{1}{2} ( - \frac{ \cos(5x) }{5} + \cos(x) )$

$= - \frac{ \cos(5x) }{10} + \frac{ \cos(x) }{2}$

Now, add the constant of integration,

$= - \frac{ \cos(5x) }{10} + \frac{ \cos(x) }{2} + C</p><p>,C \in \R$

Answer:-

• $- \frac{ \cos(5x) }{10} + \frac{ \cos(x) }{2} + C,C \in \R$