# Integrate the following trigonometric term:

$\int {\cos 3x\cos 4x.dx} $

Last updated date: 17th Mar 2023

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Hint- in order to solve such types of integral first try to simplify the terms with the help of some common trigonometric identities and then proceed with the integration part.

Complete step-by-step solution -

To find out $\int {\cos 3x\cos 4x.dx} $

By multiplying the given integral by $\dfrac{2}{2}$ , we won’t alter the integral but will bring it to satisfy some common trigonometric identity

$\dfrac{2}{2} \times \int {\cos 3x\cos 4x.dx} = \dfrac{1}{2}\int {2\cos 3x\cos 4x.dx} $

Now, first solving the internal part of the integral.

As we know the trigonometric identity

$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$

Using the above formula in the integral we get

$

\Rightarrow \dfrac{1}{2}\int {2\cos 3x\cos 4x.dx} \\

\Rightarrow \dfrac{1}{2}\int {\left[ {\cos \left( {4x + 3x} \right) + \cos \left( {4x - 3x} \right)} \right]dx} \\

\Rightarrow \dfrac{1}{2}\int {\left[ {\cos 7x + \cos x} \right]dx} \\

\Rightarrow \dfrac{1}{2}\int {\cos 7x.dx} + \dfrac{1}{2}\int {\cos x.dx} \\

$

Now as we know the formula of integral for cosine which is

$\int {\cos \theta .d\theta } = \sin \theta $

So using the same formula, we get the value of the integral which is:

$

\Rightarrow \dfrac{1}{2}\int {\cos 7x.dx} + \dfrac{1}{2}\int {\cos x.dx} \\

= \dfrac{1}{2}\dfrac{{\sin 7x}}{7} + \dfrac{1}{2}\sin x \\

= \dfrac{{\sin 7x}}{{14}} + \dfrac{{\sin x}}{2} \\

$

Hence, the value of the given integral is $\dfrac{{\sin 7x}}{{14}} + \dfrac{{\sin x}}{2}$ .

Note- In order to solve such questions related to integral of complex trigonometric identity, always try to simplify the terms in the integral part before moving to the integral part of the solution. The trigonometric identity mentioned along with the solution which helps in simplification of terms is very useful and must be remembered. Also remember the formulas for integration of some general terms.

Complete step-by-step solution -

To find out $\int {\cos 3x\cos 4x.dx} $

By multiplying the given integral by $\dfrac{2}{2}$ , we won’t alter the integral but will bring it to satisfy some common trigonometric identity

$\dfrac{2}{2} \times \int {\cos 3x\cos 4x.dx} = \dfrac{1}{2}\int {2\cos 3x\cos 4x.dx} $

Now, first solving the internal part of the integral.

As we know the trigonometric identity

$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$

Using the above formula in the integral we get

$

\Rightarrow \dfrac{1}{2}\int {2\cos 3x\cos 4x.dx} \\

\Rightarrow \dfrac{1}{2}\int {\left[ {\cos \left( {4x + 3x} \right) + \cos \left( {4x - 3x} \right)} \right]dx} \\

\Rightarrow \dfrac{1}{2}\int {\left[ {\cos 7x + \cos x} \right]dx} \\

\Rightarrow \dfrac{1}{2}\int {\cos 7x.dx} + \dfrac{1}{2}\int {\cos x.dx} \\

$

Now as we know the formula of integral for cosine which is

$\int {\cos \theta .d\theta } = \sin \theta $

So using the same formula, we get the value of the integral which is:

$

\Rightarrow \dfrac{1}{2}\int {\cos 7x.dx} + \dfrac{1}{2}\int {\cos x.dx} \\

= \dfrac{1}{2}\dfrac{{\sin 7x}}{7} + \dfrac{1}{2}\sin x \\

= \dfrac{{\sin 7x}}{{14}} + \dfrac{{\sin x}}{2} \\

$

Hence, the value of the given integral is $\dfrac{{\sin 7x}}{{14}} + \dfrac{{\sin x}}{2}$ .

Note- In order to solve such questions related to integral of complex trigonometric identity, always try to simplify the terms in the integral part before moving to the integral part of the solution. The trigonometric identity mentioned along with the solution which helps in simplification of terms is very useful and must be remembered. Also remember the formulas for integration of some general terms.

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