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Question: Integrate the given \[\int {({{\sin }^4}x - {{\cos }^4}x)dx} \] is equal to A.\[ - \dfrac{{\cos 2x...

Integrate the given ∫(sin⁡4x−cos⁡4x)dx\int {({{\sin }^4}x - {{\cos }^4}x)dx} is equal to
A.−cos⁡2x2+C - \dfrac{{\cos 2x}}{2} + C
B.−sin⁡2x2+C - \dfrac{{\sin 2x}}{2} + C
C.sin⁡2x2+C\dfrac{{\sin 2x}}{2} + C
D.cos⁡2x2+C\dfrac{{\cos 2x}}{2} + C

Explanation

Solution

We’ll be converting the cos⁡4x{\cos ^4}x or (cos⁡2x)2{({\cos ^2}x)^2} to (1−sin⁡2x)2{(1 - {\sin ^2}x)^2} and then will use this basic trigonometric identity to ultimately solve the question. We are also going to use some elementary expansion: (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab.

Formula Used:
We’ll use a basic trigonometric formula:
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
or,
cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x
And we’ll also use an elementary ”expansion of a square” formula:
(a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab

Complete step-by-step answer:
∫(sin⁡4x−cos⁡4x)dx\int {({{\sin }^4}x - {{\cos }^4}x)dx}
Putting in the above trigonometric formula,
=∫(sin⁡4x−(1−sin⁡2x)2)dx\int {({{\sin }^4}x - {{(1 - {{\sin }^2}x)}^2})dx}
Putting in the above square formula,
=∫(sin⁡4x−(1+sin⁡4x−2sin⁡2x))dx\int {({{\sin }^4}x - (1 + {{\sin }^4}x - 2{{\sin }^2}x))dx}
Opening the brackets,
=∫(sin⁡4x−1−sin⁡4x+2sin⁡2x)dx\int {({{\sin }^4}x - 1 - {{\sin }^4}x + 2{{\sin }^2}x)dx}
Cancelling out the sin⁡4x{\sin ^4}xwith the −sin⁡4x - {\sin ^4}x
=∫(−1+2sin⁡2x)dx\int {( - 1 + 2{{\sin }^2}x)dx}
Taking the negative sign common,
=∫−(1−2sin⁡2x)dx\int { - \left( {1 - 2{{\sin }^2}x} \right)dx}
Taking the negative sign out of the integral as it does not affect the result and helps in the tidiness of the integral and hence, it ultimately helps in solving the integral more effectively.
=−∫(1−2sin⁡2x)dx- \int {\left( {1 - 2{{\sin }^2}x} \right)dx}
=−∫cos⁡2xdx- \int {\cos 2xdx}
Now, we just have the answer, but we have to divide the whole by the derivative of the function’s argument’s (or coefficient’s) derivative:
=I(integral)=I({\rm{integral}}) = integral of the function (here cos)/derivative of its coefficient/argument (here 2x)
Integral of cos⁡2x=sin⁡2x\cos 2x = \sin 2x
and, derivative of its coefficient (2x) = d(2x)dx=2dxdx=2\dfrac{{d(2x)}}{{dx}} = 2\dfrac{{dx}}{{dx}} = 2
Thus,
=−∫cos⁡2xdx=−sin⁡2x2 - \int {\cos 2xdx} = - \dfrac{{\sin 2x}}{2}
and we’ll have to add an arbitrary constant as it’s an indefinite integral, so
=−∫cos⁡2xdx=−sin⁡2x2+C - \int {\cos 2xdx} = - \dfrac{{\sin 2x}}{2} + C
So, the answer of the given question is B. −sin⁡2x2+C - \dfrac{{\sin 2x}}{2} + C

Note: Here we saw how to solve a complex question by simply breaking it down into simple form and then using the basic identities. It’s a point to remember that since this is an indefinite integral, we’ve to add an arbitrary constant, in this case CC. Also, it’s a point to remember that the integral of cos⁡A\cos A is sin⁡A,\sin A, while the integral of sin⁡A\sin Ais −cos⁡A. - \cos A. Whereas the derivative (differentiation) of cos⁡A\cos A is −sin⁡A, - \sin A, while that of the sin⁡A\sin Ais cos⁡A.\cos A.