Question

plz do the Integration of ​

Answer 1

We have to find,

$\displaystyle\int\cos(3x)\cos(4x)dx$

What we have to integrate is the product of cosine ratios of two angles 3x and 4x.

$\cos(3x)\cos(4x)$

First we're going to write this product as sum of two ratios, using the product - to - sum formula.

Here it is suitable to use,

$\cos\alpha\cdot\cos\beta=\dfrac{1}{2}\bigg[\cos(\alpha-\beta)+\cos(\alpha+\beta)\bigg]$

So,

$\cos(3x)\cos(4x)=\dfrac{1}{2}\bigg[\cos(3x-4x)+\cos(3x+4x)\bigg]\\ \\ \\ =\dfrac{1}{2}\bigg[\cos(-x)+\cos(7x)\bigg]=\dfrac{1}{2}\bigg[\cos(x)+\cos(7x)\bigg]$

Now,

$\displaystyle\int\cos(3x)\cos(4x)dx\ =\ \int\dfrac{1}{2}\bigg[\cos(x)+\cos(7x)\bigg]dx\\ \\ \\ =\dfrac{1}{2}\int(\cos(x)+\cos(7x))dx\\ \\ \\ =\dfrac{1}{2}\left[\int\cos(x)dx+\int\cos(7x)dx\right]$

Consider each integral in this equation.

$\displaystyle\int\cos(x)dx$

This is generally,

$\displaystyle\int\cos(x)\dx\ =\ \sin(x)$

We know this.

What about  $\displaystyle\int\cos(7x)dx\ \ ?$

It's the integration of  cos(7x)  with respect to x. First we have to write dx with respect to 7x, i.e., we have to get it as a multiple of d(7x).

For this consider the derivative of 7x.

$\dfrac{d(7x)}{dx}\ =7\ \ \Longrightarrow\ \ dx=\dfrac{1}{7}\ d(7x)$

So,

$\displaystyle\int\cos(7x)dx\ =\ \int\dfrac{1}{7}\cos(7x)\ d(7x)\\ \\ \\ =\ \dfrac{1}{7}\int\cos(7x)\ d(7x)\ =\ \dfrac{1}{7}\sin(7x)$

Now,

$\displaystyle\dfrac{1}{2}\left[\int\cos(x)dx+\int\cos(7x)dx\right]\\ \\ \\ =\ \dfrac{1}{2}\bigg[\sin(x)+\dfrac{1}{7}\sin(7x)\bigg]\\ \\ \\ =\ \dfrac{1}{2}\sin(x)+\dfrac{1}{14}\sin(7x)$

Hence Integrated!