Question

Evaluate the following integral: $\int{\dfrac{\cos 8x-\cos 7x}{1+2\cos 5x}dx}$.

Answer 1

Hint: To evaluate the integral, we have to convert the terms in it to simpler terms using some basic trigonometric formulas and identities. Some of which we are going to need in this question are given below:
$\cos a-\cos b=-2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$
$\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$
$\sin 3x=3\sin x-4{{\sin }^{3}}x$
$2\sin A\sin B=\cos (A-B)-\cos (a+B)$

Complete step-by-step answer:
The integral given in the question is $\int{\dfrac{\cos 8x-\cos 7x}{1+2\cos 5x}dx}$.
In the first step, we will apply the identity $\cos a-\cos b=-2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ in the numerator. So, we can write $\cos 8x-\cos 7x$ as $\cos 8x-\cos 7x=-2\sin \left( \dfrac{8x+7x}{2} \right)\sin \left( \dfrac{8x-7x}{2} \right)$
$\Rightarrow \cos 8x-\cos 7x=-2\sin \left( \dfrac{15x}{2} \right)\sin \left( \dfrac{x}{2} \right)$
Now, in the second step, we are going to write $\cos 5x$ in a different form with the help of the identity $\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$. So, $\cos 5x$ can be written as $\cos 5x=1-2{{\sin }^{2}}\dfrac{5x}{2}$.
Now, we can substitute the values of the modified numerator and denominator into the given integral. After substitution, we get:
$\int{\dfrac{-2\sin \left( \dfrac{15x}{2} \right)\sin \left( \dfrac{x}{2} \right)}{1+2\left( 1-2{{\sin }^{2}}\dfrac{5x}{2} \right)}}dx$
Now, we have $\sin \left( \dfrac{15x}{2} \right)$ in the numerator. We are going to write it in the form of $\sin \dfrac{5x}{2}$ and according to identity $\sin 3x=3\sin x-4{{\sin }^{3}}x$, we can write $\sin \left( \dfrac{15x}{2} \right)=\sin \left( 3\times \dfrac{5x}{2} \right)$ as $\sin \left( \dfrac{15x}{2} \right)=3\sin \dfrac{5x}{2}-4{{\sin }^{3}}\dfrac{5x}{2}$
After substituting this value in above integral, we get:
$\int{\dfrac{-2\left( 3\sin \dfrac{5x}{2}-4{{\sin }^{3}}\dfrac{5x}{2} \right)\sin \left( \dfrac{x}{2} \right)}{1+2\left( 1-2{{\sin }^{2}}\dfrac{5x}{2} \right)}}dx$
Now, we take $\sin \dfrac{5x}{2}$ common from the numerator. After this, we get:
$\int{\dfrac{-2\sin \dfrac{5x}{2}\left( 3-4{{\sin }^{2}}\dfrac{5x}{2} \right)\sin \left( \dfrac{x}{2} \right)}{1+2\left( 1-2{{\sin }^{2}}\dfrac{5x}{2} \right)}}dx$
$\Rightarrow \int{\dfrac{-2\sin \dfrac{5x}{2}\left( 3-4{{\sin }^{2}}\dfrac{5x}{2} \right)\sin \left( \dfrac{x}{2} \right)}{3-4{{\sin }^{2}}\dfrac{5x}{2}}}dx$
Now, we cancel the common terms from numerator and denominator. After cancelling, we get: $\Rightarrow \int{-2\sin \dfrac{5x}{2}\sin \left( \dfrac{x}{2} \right)}dx$
Taking (-1) out of the integral, we get: $-\int{2\sin \dfrac{5x}{2}\sin \left( \dfrac{x}{2} \right)}dx$.
Now, with the help of identity $2\sin A\sin B=\cos (A-B)-\cos (a+B)$, we can write $2\sin \dfrac{5x}{2}\sin \left( \dfrac{x}{2} \right)$ as $2\sin \dfrac{5x}{2}\sin \dfrac{x}{2}=\cos \left( \dfrac{5x}{2}-\dfrac{x}{2} \right)-\cos \left( \dfrac{5x}{2}+\dfrac{x}{2} \right)$. Simplifying, we get, $2\sin \dfrac{5x}{2}\sin \dfrac{x}{2}=\cos \left( 2x \right)-\cos \left( 3x \right)$
We put this value in the integral and we get: $-\int{\left( \cos \left( 2x \right)-\cos \left( 3x \right) \right)}dx$. After simplification, we get: $\int{\cos 3xdx-\int{\cos 2xdx}}$
Above integral is of simple form, so we can integrate it. Therefore, we get: $\int{\cos 3xdx-\int{\cos 2xdx}}=\dfrac{\sin 3x}{3}-\dfrac{\sin 2x}{2}+c$
$\Rightarrow \int{\dfrac{\cos 8x-\cos 7x}{1+2\cos 5x}dx}=\dfrac{\sin 3x}{3}-\dfrac{\sin 2x}{2}+c$ is our required answer.

Note: The student must not forget to write + c at the end of integration since it is an indefinite integration. There is a possibility that the student might use different identities and try to simplify it. This will lead to a waste of time, so the student must first analyse the form of terms and then decide which identity to apply. The student must be careful in the signs on the identity, any change will lead to affect the answer.