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Question

Question: Integrate the given function \({\sin ^4}x\)...

Integrate the given function sin4x{\sin ^4}x

Explanation

Solution

To integrate sin4x{\sin ^4}x first we have to change the sin4x{\sin ^4}x by using cos2θ=12sin2θ\cos 2\theta = 1 - 2{\sin ^2}\theta into (1cos2x2)2{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}. It is also known that (ab)2=a2+b22ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab so (1cos2x2)2=(1+cos22x2cos2x4){\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2} = \left( {\dfrac{{1 + {{\cos }^2}2x - 2\cos 2x}}{4}} \right). After that we will convert cos22x{\cos ^2}2x by using cos2θ=2cos2θ1\cos 2\theta = 2{\cos ^2}\theta - 1 then on simplifying we get the value sin4x{\sin ^4}x will be equal to 18(3+cos4x4cos2x)\dfrac{1}{8}\left( {3 + \cos 4x - 4\cos 2x} \right). Now we can integrate this. And according to the integration theorem we can separate the addition terms,1dx=x\int 1 dx = x and cosθ=sinθ\int {\cos } \theta = \sin \theta .

Complete step-by-step answer:
(sin4x)dx\int {\left( {{{\sin }^4}x} \right)} dx
Converting sin2x{\sin ^2}x by using cos2θ=12sin2θ\cos 2\theta = 1 - 2{\sin ^2}\theta , we get
(1cos2x2)2dx\Rightarrow \int {{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}^2}} dx
Now, factorize (1cos2x)2{\left( {1 - \cos 2x} \right)^2} by using (ab)2=a2+b22ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we get
14(1+cos22x2cos2x)dx\Rightarrow \dfrac{1}{4}\int {\left( {1 + {{\cos }^2}2x - 2\cos 2x} \right)dx}
Convert cos22x{\cos ^2}2x by using cos2θ=2cos2θ1\cos 2\theta = 2{\cos ^2}\theta - 1, we get
14(12cos2x+cos4x+12)dx\Rightarrow \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + \dfrac{{\cos 4x + 1}}{2}} \right)} dx
18(3+cos4x4cos2x)dx\Rightarrow \dfrac{1}{8}\int {\left( {3 + \cos 4x - 4\cos 2x} \right)} dx
We can also write it as
38dx+18(cos4x)dx12(cos2x)dx\Rightarrow \dfrac{3}{8}\int {dx} + \dfrac{1}{8}\int {\left( {\cos 4x} \right)} dx - \dfrac{1}{2}\int {\left( {\cos 2x} \right)dx}
It is known that 1dx=x\int 1 dx = x and cosθ=sinθ\int {\cos } \theta = \sin \theta , then on simplifying, we get
3x8+sin4x32sin2x4+c\Rightarrow \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} - \dfrac{{\sin 2x}}{4} + c
(sin4x)dx=3x8+sin4x32sin2x4+c\therefore \int {\left( {{{\sin }^4}x} \right)} dx = \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} - \dfrac{{\sin 2x}}{4} + c

Note: Trigonometric formula used in this integration are cos2θ=12sin2θ\cos 2\theta = 1 - 2{\sin ^2}\theta ,cos2θ=2cos2θ1\cos 2\theta = 2{\cos ^2}\theta - 1, whereas formula of integration uses are 1dx=x\int 1 dx = x and cosθ=sinθ\int {\cos } \theta = \sin \theta . Point to take care there is no any formula of direct integration of higher power of trigonometric ratio we have to change it into power 1 by using trigonometric formula. Constant in the integration can be taken out from the integration.