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Question: How do you integrate \(\int {x\sqrt {1 - {x^4}} dx} \) using trigonometric substitution?...

How do you integrate ∫x1−x4dx\int {x\sqrt {1 - {x^4}} dx} using trigonometric substitution?

Explanation

Solution

According to the question we have to determine the integration of ∫x1−x4dx\int {x\sqrt {1 - {x^4}} dx} using trigonometric substitution. So, first of all to determine the integration of ∫x1−x4dx\int {x\sqrt {1 - {x^4}} dx} using trigonometric substitution we have to let x2=sin⁡t{x^2} = \sin t.
Now, we have to differentiate to what we have let is x2=sin⁡t{x^2} = \sin t with the help of the formula of the differentiation as mentioned below:

Formula used:
dxndx=nxn−1 dsin⁡xdx=cos⁡x  \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}} \\\ \dfrac{{d\sin x}}{{dx}} = \cos x \\\ ………………….(A)
Now, we have to substitute all the values which are obtained after the differentiation in the given function to obtain the integration.
Now, to solve the integration we have to use the identity which is as mentioned below:
Identity used:
⇒sin⁡2x+cos⁡2x=1..............(B)\Rightarrow {\sin ^2}x + {\cos ^2}x = 1..............(B)
Now, we have to use some other important identity which is as mentioned below:
⇒cos⁡2x=cos⁡2x−sin⁡2x..............(C)\Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x..............(C)
⇒cos⁡2x=2cos⁡2x−1 ⇒cos⁡2x=1+cos⁡2x2..............(D)  \Rightarrow \cos 2x = 2{\cos ^2}x - 1 \\\ \Rightarrow {\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}..............(D) \\\
Now, we have to use trigonometric functions to solve the obtained integration.
Formula used:
⇒∫1dx=x+c ⇒∫cos⁡axdx=cos⁡axa+c  \Rightarrow \int {1dx = x + c} \\\ \Rightarrow \int {\cos axdx = \dfrac{{\cos ax}}{a} + c} \\\ ……………(E)
Now, we have to use the identity which is as mentioned below to obtain the required integration of the given function.
Identity used:
⇒sin⁡2x=2sin⁡xcos⁡x......................(F)\Rightarrow \sin 2x = 2\sin x\cos x......................(F)
Now, we have to substitute the value of t as we let in the initial part of the solution to determine the actual integration of the given function.

Complete step by step solution:
Step 1: first of all to determine the integration of ∫x1−x4dx\int {x\sqrt {1 - {x^4}} dx} using trigonometric substitution we have to letx2{x^2}to some trigonometric function. Hence,
Let,
x2=sin⁡t{x^2} = \sin t
Step 2: Now, we have to differentiate to what we have let is x2=sin⁡t{x^2} = \sin twith the help of the formula (A) of the differentiation as mentioned in the solution hint. Hence,
⇒2xdx=cos⁡tdt\Rightarrow 2xdx = \cos tdt
Step 3: Now, we have to substitute all the values which are obtained after the differentiation in the given function to obtain the integration. Hence,
⇒∫121−x42xdx\Rightarrow \int {\dfrac{1}{2}\sqrt {1 - {x^4}} 2xdx}
On substituting the values in the integral as obtained just above,
⇒∫12cos⁡t1−sin⁡2tdt\Rightarrow \int {\dfrac{1}{2}\cos t\sqrt {1 - {{\sin }^2}t} dt}
Step 4: Now, to solve the integration we have to use the identity (B) which is as mentioned in the solution hint. Hence,
⇒∫12cos⁡2tdt\Rightarrow \int {\dfrac{1}{2}{{\cos }^2}tdt}
Step 5: Now, we have to use some other important identity (B) which is as mentioned in the solution hint. Hence,
⇒∫12×12×(1−cos⁡2t)dt ⇒14∫1dt+14∫cos⁡2tdt  \Rightarrow \int {\dfrac{1}{2} \times \dfrac{1}{2} \times (1 - \cos 2t)dt} \\\ \Rightarrow \dfrac{1}{4}\int {1dt + \dfrac{1}{4}\int {\cos 2t} dt} \\\
On solving the integration as obtained just above, with the help of the formula (E) as mentioned in the solution hint.

⇒14t+14sin⁡2t2+c ⇒14(t+sin⁡2t2)+c  \Rightarrow \dfrac{1}{4}t + \dfrac{1}{4}\dfrac{{\sin 2t}}{2} + c \\\ \Rightarrow \dfrac{1}{4}\left( {t + \dfrac{{\sin 2t}}{2}} \right) + c \\\

Where, c is the constant.
Step 6: Now, we have to use the identity (F) which is as mentioned in the solution hint to obtain the required integration of the give function. Hence,
⇒14(t+2sin⁡tcos⁡t2)+c ⇒14(t+sin⁡tcos⁡t)+c  \Rightarrow \dfrac{1}{4}\left( {t + \dfrac{{2\sin t\cos t}}{2}} \right) + c \\\ \Rightarrow \dfrac{1}{4}(t + \sin t\cos t) + c \\\
Now, using the identity as mentioned in the solution hint,
⇒14(t+sin⁡t1−sin⁡2t)+c\Rightarrow \dfrac{1}{4}(t + \sin t\sqrt {1 - {{\sin }^2}t} ) + c
Step 7: Now, we have to substitute the value of t as we let in the initial part of the solution to determine the actual integration of the given function. Hence,
=14(sin⁡−1(x2)+x21−x2)+c= \dfrac{1}{4}({\sin ^{ - 1}}({x^2}) + {x^2}\sqrt {1 - {x^2}} ) + c

Hence, with the help of the formulas and identities as mentioned in the solution hint we have determined the required integral which is=14(sin⁡−1(x2)+x21−x2)+c = \dfrac{1}{4}({\sin ^{ - 1}}({x^2}) + {x^2}\sqrt {1 - {x^2}} ) + c.

Note:

  1. It is necessary that we have to let x2=sin⁡t{x^2} = \sin tas asked in the question that we have to find the integration using the substitution method and then we have to integrate the term with respect to x.
  2. Don’t forget to replace the value of t as we let in the beginning of the solution and during substitution we can also take help of the identities which are mentioned in the solution hint.