Question

How do you find the integral of \[{\cos ^4}t \cdot {\sin ^2}t\] ?

Answer 1

Hint: Integration is the process of finding the antiderivative. The integration of \[g'\left( x \right)\]with respect to dx is given by \[\int {{g^1}\left( x \right)dx = g\left( x \right) + C} \].Here C is the constant of integration and we can find the integral by using trigonometric identities of cos and sine functions and finding the integral of terms.

Complete step by step answer:
The given function is \[{\cos ^4}t \cdot {\sin ^2}t\].
As we need to find the integral, let us rewrite the function as
\[I = \int {{{\cos }^4}t \cdot {{\sin }^2}t \cdot dt} \] ………………… 1
Observe that, cos and sin both have even Power. So, we have to convert them into Multiple Angles, using the Identities as:
 \[{\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}\]
\[\Rightarrow{\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}\]
\[\Rightarrow 2\cos \alpha \cos \beta = \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)\]
Now let us apply the identities with respect to the function given as
\[{\cos ^4}t \cdot {\sin ^2}t = \dfrac{1}{4}\left( {4{{\cos }^2}t \cdot {{\sin }^2}t} \right)\left( {{{\cos }^2}t} \right)\]
Simplifying the functions, we get
\[\dfrac{1}{4}\left( {{{\sin }^2}\left( {2t} \right)} \right)\left( {{{\cos }^2}t} \right)\]
Expanding the sin and cos function as
\[\dfrac{1}{4}\left\{ {\dfrac{1}{2}\left( {1 - \cos 4t} \right)} \right\}\left\{ {\dfrac{1}{2}\left( {1 + \cos 2t} \right)} \right\}\]
\[\Rightarrow\dfrac{1}{{16}}\left( {1 - \cos 4t + \cos 2t - \cos 4t\cos 2t} \right)\]
\[\Rightarrow\dfrac{1}{{32}}\left\{ {2 - 2\cos 4t + 2\cos 2t - 2\cos 4t\cos 2t} \right\}\]
\[\Rightarrow\dfrac{1}{{32}}\left\{ {2 - 2\cos 4t + 2\cos 2t - \left( {\cos 6t + \cos 2t} \right)} \right\}\]
\[\Rightarrow\dfrac{1}{{32}}\left\{ {2 - \cos 6t - 2\cos 4t + 3\cos 2t} \right\}\] …………… 2
Therefore, to find the integral by equation 1
\[I = \int {{{\cos }^4}t \cdot {{\sin }^2}t \cdot dt} \]
Substituting the solution of \[{\cos ^4}t \cdot {\sin ^2}t\] i.e., from equation 2 we can find the integral
\[I = \int {\dfrac{1}{{32}}\left\{ {2 - \cos 6t - 2\cos 4t + 3\cos 2t} \right\}dt} \]
Integrate the terms of the function as
\[\dfrac{1}{{32}}\left( {2t - \dfrac{1}{6}\sin 6t - 2 \cdot \dfrac{1}{4}\sin 4t + 3 \cdot \dfrac{1}{2}\sin 2t} \right)\]
Therefore, after integration we get
\[\therefore I = \dfrac{1}{{192}}\left( {12t - \sin 6t - 3\sin 4t + 9\sin 2t} \right) + C\]

Hence,the integral of \[{\cos ^4}t \cdot {\sin ^2}t\] is $ \dfrac{1}{{192}}\left( {12t - \sin 6t - 3\sin 4t + 9\sin 2t} \right) + C$.

Note:There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. Hence, based on the function given we can find the integration of the function i.e., by using the integration methods as the details are given as additional information.