We have an answer to "In the following exercises, find each indefinite integral,..."

hadejada7x

Answered question

2021-12-13

In the following exercises, find each indefinite integral, using appropriate substitutions.

\int \frac{d x}{\sqrt{1 - 16 x^{2}}}

Vivian Soares

Beginner2021-12-14Added 36 answers

Step 1
We need to evaluate the following integral

I = \int \frac{d x}{\sqrt{1 - 16 x^{2}}}

...(1)
Step 2
Let

x = \frac{\sin u}{4} \Rightarrow d x = \frac{\cos u}{4} d u

and substitute these in equation (1). We get,

I = \int \frac{1}{\sqrt{1 - 16 {(\frac{\sin u}{4})}^{2}}} (\frac{\cos u}{4} d u)

= \frac{1}{4} \int \frac{\cos u}{\sqrt{1 - \sin^{2} u}} d u

= \frac{1}{4} \int \frac{\cos u}{\sqrt{\cos^{2} u}} d u ∵ 1 - \sin^{2} u = \cos^{2} u

= \frac{1}{4} \int \frac{\cos u}{\cos u} d u

= \frac{1}{4} \int d u

= \frac{1}{4} u + C

= \frac{1}{4} \sin^{- 1} (4 x) + C ∵ x = \frac{\sin u}{4} \Rightarrow u = \sin^{- 1} (4 x)

Hence, the integral is

\int \frac{d x}{\sqrt{1 - 16 x^{2}}} = \frac{1}{4} \sin^{- 1} (4 x) + C

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