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Question: Evaluate the following integral \(\int{\dfrac{\cos 3x}{\cos x}}\)....

Evaluate the following integral ∫cos⁡3xcos⁡x\int{\dfrac{\cos 3x}{\cos x}}.

Explanation

Solution

In this question, we need to find the integration of a function f(x)=cos⁡3xcos⁡xf\left( x \right)=\dfrac{\cos 3x}{\cos x}. For this, we will first simplify f(x) to find the integration. We will use the formula cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4{{\cos }^{3}}x-3\cos x and cos⁡2x=2cos⁡2x−1\cos 2x=2{{\cos }^{2}}x-1 to simplify the function f(x).
After that, we will use the basic integration formula of cosax given by ∫cos⁡axdx=sin⁡axa\int{\cos axdx}=\dfrac{\sin ax}{a}. We will also use ∫1⋅dx=x\int{1\cdot dx=x}

Complete step-by-step answer:
Here, we need to find the integration of the function f(x)=cos⁡3xcos⁡xf\left( x \right)=\dfrac{\cos 3x}{\cos x}. But first let us simplify it.
As we know, cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4{{\cos }^{3}}x-3\cos x so our function f(x) becomes f(x)=4cos⁡3x−3cos⁡xcos⁡xf\left( x \right)=\dfrac{4{{\cos }^{3}}x-3\cos x}{\cos x}.
Separating terms we get: f(x)=4cos⁡3xcos⁡x−3cos⁡xcos⁡xf\left( x \right)=\dfrac{4{{\cos }^{3}}x}{\cos x}-\dfrac{3\cos x}{\cos x}.
Cancelling cos x from the numerator and denominator from both terms we get: f(x)=4cos⁡2x−3f\left( x \right)=4{{\cos }^{2}}x-3.
Now, we know that, cos⁡2x=2cos⁡2x−1\cos 2x=2{{\cos }^{2}}x-1 so we get: cos⁡2x+1=2cos⁡2x\cos 2x+1=2{{\cos }^{2}}x.
Putting it in f(x) we get: f(x)=4(cos⁡2x+1)2−3f\left( x \right)=\dfrac{4\left( \cos 2x+1 \right)}{2}-3.
Simplifying it we get: f(x)=2(cos⁡2x+1)−3=2cos⁡2x+2−3=2cos⁡2x−1f\left( x \right)=2\left( \cos 2x+1 \right)-3=2\cos 2x+2-3=2\cos 2x-1.
Now, we need to find the integration of f(x). Therefore, ∫f(x)dx=∫(2cos⁡2x−1)dx\int{f\left( x \right)}dx=\int{\left( 2\cos 2x-1 \right)dx}.
As we know, ∫[u(x)+v(x)]dx=∫u(x)dx+∫v(x)dx\int{\left[ u\left( x \right)+v\left( x \right) \right]}dx=\int{u\left( x \right)dx+\int{v\left( x \right)dx}} so we get: ∫f(x)dx=∫2cos⁡2xdx+∫(−1)dx\int{f\left( x \right)dx=\int{2\cos 2xdx+\int{\left( -1 \right)dx}}}.
Also we know that ∫ag(x)dx=a∫g(x)dx\int{ag\left( x \right)}dx=a\int{g\left( x \right)}dx hence we get: ∫f(x)dx=2∫cos⁡2xdx−∫dx\int{f\left( x \right)}dx=2\int{\cos 2xdx-\int{dx}}.
Using ∫cos⁡axdx=sin⁡axa and ∫dx=x\int{\cos axdx=\dfrac{\sin ax}{a}}\text{ and }\int{dx}=x we get: ∫f(x)dx=2(sin⁡2x2)−x\int{f\left( x \right)}dx=2\left( \dfrac{\sin 2x}{2} \right)-x.
Cancelling 2 we get: ∫f(x)dx=sin⁡2x−x\int{f\left( x \right)}dx=\sin 2x-x
As f(x) was supposed to be cos⁡3xcos⁡x\dfrac{\cos 3x}{\cos x} so we get: ∫cos⁡3xcos⁡x=sin⁡2x−x\int{\dfrac{\cos 3x}{\cos x}=\sin 2x-x} which is our required answer.

Note: Students should keep in mind the formula of finding integration of the trigonometric function. Students should note that we always need to simplify our function as there exists no formula for finding integration of the two dividing functions.