Question

How do you find the integral $\int {x \cdot \sqrt {1 - {x^4}} } dx$ ?

Answer 1

Hint: Here we integrate the numerical by following the substitution method.
On replace ${x^2}$ with $\sin t$. We differentiate both sides of ${x^2} = \sin t$ and get our answer as $2xdx = \cos tdt$.
We replace \[2xdx\] with \[\dfrac{1}{2}\cos tdt\]and then solve our integral further using trigonometric identities.
After we have simplified our sum, we just simply integrate the terms. Then we place back the original variables in the substituted variables to get our required answer.

Formula used: ${\sin ^2}t + {\cos ^2}t = 1$, $\cos 2t = {\cos ^2}t - {\sin ^2}t$ and $\sin 2t = 2\sin t\cos t$

Complete step-by-step solution:
The given integral is $\int {x \cdot \sqrt {1 - {x^4}} } dx$.
We can also write the given integral in the form: $\int {x \cdot \sqrt {1 - {{\left( {{x^2}} \right)}^2}} } dx$.…… let’s call this as equation (A)
Now, In order to solve this, we use the substitution method to make solving our sum easier.
Let’s substitute ${x^2} = \sin t$
Now on differentiating both the sides of the above equation, we get:
$2xdx = \cos tdt$. Placing this in equation (A), we get:
$ \Rightarrow \int {\dfrac{1}{2}\sqrt {1 - {{\sin }^2}tdt} \left( {2xdx} \right)} $ ………….we take $\dfrac{1}{2}$ outside of the bracket in order to balance the $2xdx$
We know that $2xdx = \cos tdt$, therefore our equation becomes:
$ \Rightarrow \int {\dfrac{1}{2}\cos tdt\sqrt {1 - {{\sin }^2}tdt} } $
Now according to the identity: ${\sin ^2}t + {\cos ^2}t = 1$
Therefore, ${\cos ^2}t = 1 - {\sin ^2}t$
Placing this in our integral equation, we get:
$ \Rightarrow \int {\dfrac{1}{2}\cos tdt\sqrt {{{\cos }^2}tdt} } $
On squaring we get
$ \Rightarrow \int {\dfrac{1}{2}\cos tdt \times \cos tdt} $
On multiply the term and we get
$ \Rightarrow \int {\dfrac{1}{2}{{\cos }^2}tdt} $………………………….consider this to be equation (B)
Now according to the identity: $\cos 2t = {\cos ^2}t - {\sin ^2}t$
Now we know that ${\sin ^2}t = 1 - {\cos ^2}t$
Therefore, $\cos 2t = {\cos ^2}t - \left( {1 - {{\cos }^2}t} \right)$
On expanding it, we get:
$ \Rightarrow \cos 2t = {\cos ^2}t - 1 + {\cos ^2}t = 2{\cos ^2}t - 1$
Thus, $\cos 2t = 2{\cos ^2}t - 1$
Now adding $ + 1$ to both sides, we get:
$ \Rightarrow \cos 2t + 1 = 2{\cos ^2}t$
Dividing both sides with$2$ , we get:
$ \Rightarrow \dfrac{{\cos 2t + 1}}{2} = {\cos ^2}t$
On placing the value of ${\cos ^2}t$ in equation (B) which is $\int {\dfrac{1}{2}{{\cos }^2}tdt} $, we get:
$ \Rightarrow \int {\dfrac{1}{2}\left( {\dfrac{{\cos 2t + 1}}{2}} \right)dt} $
Taking $\dfrac{1}{2}$ as a common factor outside, we get:
$ \Rightarrow \int {\dfrac{1}{2} \times \dfrac{1}{2}\left( {\cos 2t + 1} \right)dt} $
$ \Rightarrow \int {\dfrac{1}{4}\left( {\cos 2t + 1} \right)dt} $
Now let’s integrate within the bracket:
$ \Rightarrow \dfrac{1}{4}\int {\cos 2tdt + \int {1dt} } $
We know that $\int {\cos 2tdt = \dfrac{{\sin 2t}}{2}} $ and $\int {1dt = t} $
Therefore, we have $\dfrac{1}{4}\left[ {\dfrac{{\sin 2t}}{2} + t} \right] + C$
$\sin 2t = 2\sin t\cos t$ ⟶ placing this in the above equation, we get:
$ \Rightarrow \dfrac{1}{4}\left[ {\dfrac{{2\sin t\cos t}}{2} + t} \right] + C$
On solving it further, we get:
$ \Rightarrow \dfrac{1}{4}\left[ {\dfrac{{\not{2}\sin t\cos t}}{{\not{2}}} + t} \right] + C$
On rewriting we get
$ \Rightarrow \dfrac{1}{4}\left[ {\sin t\cos t + t} \right] + C$
We know that $\cos t = \sqrt {1 - {{\sin }^2}t} $
Therefore, $\dfrac{1}{4}\left[ {\sin t\sqrt {1 - {{\sin }^2}t} + t} \right] + C$
As we have taken ${x^2} = \sin t$
Therefore, $t = {\sin ^{ - 1}}\left( {{x^2}} \right)$
$ \Rightarrow \dfrac{1}{4}\left[ {{x^2}\sqrt {1 - {x^4}} + {{\sin }^{ - 1}}\left( {{x^2}} \right)} \right] + C$

The required answer is $\dfrac{1}{4}\left[ {{x^2}\sqrt {1 - {x^4}} + {{\sin }^{ - 1}}\left( {{x^2}} \right)} \right] + C$

Note: Integration is a type of calculus along with differentiation. Integration simply means to add up smaller parts of any area, volume, etc to give us the whole value. There are different types of integration methods to solve more complex multiplication and division questions like:
Integration by substitution
Integration by parts, etc