If {{1-x^{4}}}{x^{7}} \, dx = f(x) ≤ft( {1-x^{4}} )ⁿ + c, then (f(x))ⁿ = ______

Correct Answer is : −1216x18-{1}{216 x¹8}}−216x181

Correct Answer is : −1216x18-{1}{216 x¹8}}−216x181

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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

\int \frac{\sqrt{1-x^{4}}}{x^{7}} \, dx = f(x) \left( \sqrt{1-x^{4}} \right)^{n} + c

(f(x))^{n} =

______

Solution:

I = \int \frac{\sqrt{1-x^{4}}}{x^{7}} \, dx

Let

x^{2} = u

, then

2x \, dx = du

I = \int \frac{\sqrt{1-u^{2}}}{2 u^{4}} \, du

= \frac{1}{2} \int \frac{\sqrt{1-u^{2}}}{u^{4}} \, du

Let

u = \sin v

, then

du = \cos v \, dv

I = \frac{1}{2} \int \frac{\cos^{2} v}{\sin^{4} v} \, dv

= \frac{1}{2} \int \frac{1}{\cos^{2} v \tan^{4} v} \, dv

= \frac{1}{2} \int \frac{\sec^{2} v}{\tan^{4} v} \, dv

Let

\tan v = w

, then

\sec^{2} v \, dv = dw

I = \frac{1}{2} \int \frac{dw}{w^{4}} = \frac{1}{2} \left( \frac{w^{-3}}{-3} \right)

= -\frac{1}{6 w^{3}} + C

= -\frac{1}{6 \tan^{3} v}

= -\frac{(\sqrt{1-u^{2}})^{3}}{6 u^{3}}

= -\frac{1}{6} \frac{(1-x^{4})^{3/2}}{x^{6}} + C

\therefore I = -\frac{1}{6} \frac{(1-x^{4})^{3/2}}{x^{6}} + C

Now,

I = f(x) (1-x^{4})^{n/2} + C

on Comparing, we obtain

f(x) = -\frac{1}{6 x^{6}}

and

n = 3

\therefore (f(x))^{n} = \left( -\frac{1}{6 x^{6}} \right)^{3} = -\frac{1}{216 x^{18}}

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