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Question

Question: How do you find the integral of \( {\cos ^2}(2x)dx \) ?...

How do you find the integral of cos2(2x)dx{\cos ^2}(2x)dx ?

Explanation

Solution

Hint : Here in this question, we have to find the integral for the function, the function is of trigonometry function. First, we simplify the trigonometric function and then we apply the integration to function and we obtain the required solution for the question.

Complete step-by-step answer :
In this question we have to find the integral of the function, the function is trigonometry. The cosine is one of the trigonometry ratios.
Now consider the given function cos2(2x){\cos ^2}(2x)
To solve the above equation, we use a double angle formula. In general, the double angle for cos(2x)\cos (2x) is defined as cos(2x)=2cos2x1\cos (2x) = 2{\cos ^2}x - 1
The above can be written as cos(2x)+12=cos2x\dfrac{{\cos (2x) + 1}}{2} = {\cos ^2}x
Here in the question the value of x is 2x.
So, by using the above formula the given equation can be written as
cos2(2x)=1+cos(2(2x))2\Rightarrow {\cos ^2}(2x) = \dfrac{{1 + \cos (2(2x))}}{2}
On simplification we get
cos2(2x)=1+cos(4x)2\Rightarrow {\cos ^2}(2x) = \dfrac{{1 + \cos (4x)}}{2}
After simplifying we apply the integration, here integral is indefinite integral because we don’t have any limits points. If we have limit points then it is a definite integral.
Now we apply the integration to the above function we get
cos2(2x)dx=1+cos(4x)2dx\int {{{\cos }^2}(2x)dx} = \int {\dfrac{{1 + \cos (4x)}}{2}} dx
Let we take 12\dfrac{1}{2} outside the integral and it is common for both the terms
cos2(2x)dx=12dx+12cos(4x)dx\Rightarrow \int {{{\cos }^2}(2x)dx} = \dfrac{1}{2}\int {dx} + \dfrac{1}{2}\int {\cos (4x)dx}
The integration of dxdx is x and the integration of cos(nx)\cos (nx) is sin((n+1)x)(n+1)\dfrac{{\sin ((n + 1)x)}}{{(n + 1)}} .
By applying the integration we have
cos2(2x)dx=12(x)+12sin(4x)4+c\Rightarrow \int {{{\cos }^2}(2x)dx} = \dfrac{1}{2}(x) + \dfrac{1}{2}\dfrac{{\sin (4x)}}{4} + c
On further simplification we get
cos2(2x)dx=x2+sin(4x)8+c\Rightarrow \int {{{\cos }^2}(2x)dx} = \dfrac{x}{2} + \dfrac{{\sin (4x)}}{8} + c
Where c is the integration constant.
So, the correct answer is “ x2+sin(4x)8+c\dfrac{x}{2} + \dfrac{{\sin (4x)}}{8} + c ”.

Note : For the trigonometry ratios we have double angle formula and half angle formula. By using these formulas, we can solve the trigonometry ratios. The double angle formula for cosine is defined as cos(2x)=2cos2x1\cos (2x) = 2{\cos ^2}x - 1 and the half angle formula is defined as cos(x2)=±1+cosx2\cos \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos x}}{2}} where x represents the angle. The integration is an inverse or reciprocal of differentiation. The integral is of indefinite integral where we don’t have limit points.