Question

integrate sin2x . cos 3x .dx

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$I=\int sin2xcos3x \, dx$

Multiplying and dividing by $\frac{1}{2}$ on the RHS of the above equation, we get

$I=\frac{1}{2}\int 2sin2xcos3x \, dx$

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

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

$I=\frac{1}{2}\int sin5x\, dx-\frac{1}{2}\int sinx\, dx$

$I=-\frac{1}{10}cosx+\frac{1}{2}cosx+c$

$I=\frac{1}{2}cosx-\frac{1}{10}cosx+c$

which is the required simplified form.