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Question

Question: How do you find the integral of \({\sin ^2}\left( {5x} \right)\) ?...

How do you find the integral of sin2(5x){\sin ^2}\left( {5x} \right) ?

Explanation

Solution

In this question, we are given an equation and we have been asked to find its integral. Convert the value sin2x{\sin ^2}x into the terms of cosx\cos x using the trigonometric identity of double angle of cosx\cos x. Then, integrate the terms with respect to x. After you have integrated the entire term, add the constant of integration to the final answer.

Formula used: 1) cosaxdx=sinaxa+C\int {\cos axdx = } \dfrac{{\sin ax}}{a} + C
2) cos2x=12sin2x\cos 2x = 1 - 2{\sin ^2}x

Complete step-by-step solution:
We are given a trigonometric expression and we have to integrate it. Since the given expression is in terms of x, we will integrate it in terms of x.
sin2(5x)\Rightarrow {\sin ^2}\left( {5x} \right)
We have a direct integration of sinx\sin x, but we do not have any direct integration of sin2x{\sin ^2}x. Therefore, we will convert the equation into the terms of cosx\cos x using the trigonometric formula.
We know thatcos2x=12sin2x\cos 2x = 1 - 2{\sin ^2}x. By shifting the terms of the formula, we will get the value of sin2x{\sin ^2}x.
sin2x=1cos2x2\Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}
On splitting the term and we get
sin2x=12cos2x2\Rightarrow {\sin ^2}x = \dfrac{1}{2} - \dfrac{{\cos 2x}}{2}
But we are given sin2(5x){\sin ^2}\left( {5x} \right). Putting it in the above formula,
sin2(5x)=12cos10x2\Rightarrow {\sin ^2}\left( {5x} \right) = \dfrac{1}{2} - \dfrac{{\cos 10x}}{2}
Therefore, now we have to integrate 12cos10x2\dfrac{1}{2} - \dfrac{{\cos 10x}}{2} with respect to x.
(12cos10x2)dx\Rightarrow \int {\left( {\dfrac{1}{2} - \dfrac{{\cos 10x}}{2}} \right)} dx
Integrating with respect to x,
12dx12cos10xdx\Rightarrow \dfrac{1}{2}\int {dx - \dfrac{1}{2}\int {\cos 10x} dx}
On integrating, we will get,
x2sin10x10+C\Rightarrow \dfrac{x}{2} - \dfrac{{\sin 10x}}{{10}} + C …. (Using cosaxdx=sinaxa+C\int {\cos axdx = } \dfrac{{\sin ax}}{a} + C)

Therefore, integration of sin2(5x){\sin ^2}\left( {5x} \right) is x2sin10x10+C\dfrac{x}{2} - \dfrac{{\sin 10x}}{{10}} + C

Note: We know that integration of sinx\sin x is cosx\cos x. But what if the x of these trigonometric ratios has a coefficient? When we face this issue in differentiation, we use chain rule and multiply the ratio with that coefficient. But while integrating, what is to be done?
When we face a similar issue in integration, we divide the answer by that coefficient. It is similar to what we did in the last step of the solution. But, let us take another example. What is the integration of cos7x\cos 7x?
cos7xdx=sin7x7+C\int {\cos 7xdx = \dfrac{{ - \sin 7x}}{7} + C}
In this example, the coefficient of cosx\cos x was 7, so we divided our answer by 7.