Question
Question: Integrate the given function \({\sin ^4}x\)...
Integrate the given function sin4x
Solution
To integrate sin4x first we have to change the sin4x by using cos2θ=1−2sin2θ into (21−cos2x)2. It is also known that (a−b)2=a2+b2−2ab so (21−cos2x)2=(41+cos22x−2cos2x). After that we will convert cos22x by using cos2θ=2cos2θ−1 then on simplifying we get the value sin4x will be equal to 81(3+cos4x−4cos2x). Now we can integrate this. And according to the integration theorem we can separate the addition terms,∫1dx=x and ∫cosθ=sinθ.
Complete step-by-step answer:
∫(sin4x)dx
Converting sin2x by using cos2θ=1−2sin2θ, we get
⇒∫(21−cos2x)2dx
Now, factorize (1−cos2x)2 by using (a−b)2=a2+b2−2ab, we get
⇒41∫(1+cos22x−2cos2x)dx
Convert cos22x by using cos2θ=2cos2θ−1, we get
⇒41∫(1−2cos2x+2cos4x+1)dx
⇒81∫(3+cos4x−4cos2x)dx
We can also write it as
⇒83∫dx+81∫(cos4x)dx−21∫(cos2x)dx
It is known that ∫1dx=x and ∫cosθ=sinθ, then on simplifying, we get
⇒83x+32sin4x−4sin2x+c
∴∫(sin4x)dx=83x+32sin4x−4sin2x+c
Note: Trigonometric formula used in this integration are cos2θ=1−2sin2θ,cos2θ=2cos2θ−1, whereas formula of integration uses are ∫1dx=x and ∫cosθ=sinθ. 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.