Question

Integrate sin3x.cos4x

Answer 1

Hint- To integrate some mathematical functions like sin3x.cos4x , we first need to change the multiplication sign between them using the trigonometric formula of $\sin a.\cos b = \dfrac{1}{2}\left( {\sin \left( {a + b} \right) + \sin \left( {a - b} \right)} \right)$ and then integrate them separately using integration formulas and put a constant at the end to get the required solution.

Complete step by step answer:

We know that
Sin(a+b) = sina.cosb + cosa.sinb
Sin(a-b) = sina.cosb - cosa.sinb
On adding both the equations , we get
Sin(a+b) + sin(a-b)= 2sina.cosb
$\Rightarrow \sin a.\cos b = \dfrac{1}{2}\left( {\sin \left( {a + b} \right) + \sin \left( {a - b} \right)} \right)$
Therefore ,
sin3x.cos4x = $\dfrac{1}{2}\left( {\sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right)} \right)$
$\Rightarrow sin3x.cos4x = \dfrac{1}{2}\left( {\sin 7x - \sin x} \right){\text{ }}\left( {{\text{since sin}}\left( { - \theta } \right) = - \sin \theta } \right){\text{ }}$
Now integrating the above ,
$\int {\left( {\sin 3x.cos4x} \right)dx} = \dfrac{1}{2}\left[ {\int {\sin 7xdx} - \int {\sin xdx} } \right]$
$\Rightarrow \int {\left( {\sin 3x.cos4x} \right)dx} = \dfrac{1}{2}\left[ { - \dfrac{{\cos 7x}}{7} + \cos x} \right] + C{\text{ }}\left( {\operatorname{si} {\text{nce }}\int {\sin \left( {ax + b} \right)dx = - \dfrac{{\cos \left( {ax + b} \right)}}{a}} {\text{ }}} \right)$
$\Rightarrow \int {\left( {\sin 3x.cos4x} \right)dx} = \dfrac{{ - 1}}{{14}}\cos 7x + \dfrac{1}{2}\cos x + C$

Note - In order to include all antiderivatives of f(x) the constant of integration C is used for indefinite integrals. The importance of C is that it allows us to express the general form of antiderivatives.