Question

Trigonometric Substitution Suppose \(u=\sin ^{-1} x .\) Then \(\cos u>0\) . (a) Use the substitution \(x=\sin u, d x=\cos u d u\) to show that $$\int \frac{d x}{\sqrt{1-x^{2}}}=\int 1 d u$$ (b) Evaluate \(\int 1 d u\) to show that \(\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x+C\)

Short answer

The integral \(\int \frac{dx}{\sqrt{1-x^2}}\) is equal to \(\sin^{-1} x +C\).

Step-by-step solution

Initialize the substitution

We are given the substitution \(x=\sin u\) which implies \(dx=\cos u du\), so we substitute this in the given integral. The integral \(\int \frac{dx}{\sqrt{1-x^2}}\) becomes \(\int \frac{\cos u du}{\sqrt{1-\sin^2 u}}\).

Simplify the integral

The expression in the denominator, \(\sqrt{1-\sin^2 u}\), can be simplified using Pythagorean trigonometric identity, that is \(\sqrt{1-\sin^2 u} = \cos u\). After this simplification, the integral becomes \(\int \frac{\cos u \, du}{\cos u}=\int 1 \, du\) thereby proving the first part of the problem.

Integrate \(1 du\)

Now, we integrate \(1 du\) which gives us \(u + C\), where \(C\) is the constant of integration.

Substitute \(u\)

Now, we substitute \(u = \sin ^{-1}x\) back into the solution, which yields \(\sin^{-1}x + C\). Hence, this shows that \(\int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x + C\).