Question

rewrite the given integral using this substitution.
\frac{5}{sqrt{2}}int_{0}^{}\frac{dx}{sqrt{25 - x^{2}}}=int_{0}^{}(\text{ })\text{ }d\theta
(type exact answers.)

Explanation:

Step1: Find the derivative of \(x\)

Given \(x = 5\sec\theta\), then \(dx=5\sec\theta\tan\theta d\theta\).

Step2: Rewrite the lower - limit

When \(x = 0\), we have \(0 = 5\sec\theta\), \(\sec\theta=0\) (not possible). But from \(x = 5\sec\theta\), we know \(\cos\theta=\frac{5}{x}\). When \(x = 0\), we consider the substitution in the context of the integral form. Another way, we know the standard trig - substitution for \(\sqrt{a^{2}-x^{2}}\) is \(x=a\sin t\), but here using \(x = 5\sec\theta\), we first rewrite the integrand. The integrand is \(\frac{dx}{\sqrt{25 - x^{2}}}\), substituting \(x = 5\sec\theta\) gives \(\frac{5\sec\theta\tan\theta d\theta}{\sqrt{25-25\sec^{2}\theta}}=\frac{5\sec\theta\tan\theta d\theta}{5\sqrt{1 - \sec^{2}\theta}}\). Since \(1+\tan^{2}\theta=\sec^{2}\theta\), then \(1 - \sec^{2}\theta=-\tan^{2}\theta\). So \(\frac{5\sec\theta\tan\theta d\theta}{5\sqrt{1 - \sec^{2}\theta}}=\frac{5\sec\theta\tan\theta d\theta}{5i\tan\theta}= - i\sec\theta d\theta\) (this is wrong substitution, we should use \(x = 5\sin\theta\)). Let's start over with \(x = 5\sin\theta\), then \(dx = 5\cos\theta d\theta\).
When \(x = 0\), \(0=5\sin\theta\), so \(\theta = 0\).

Step3: Rewrite the upper - limit

When \(x=\frac{5}{\sqrt{2}}\), we have \(\frac{5}{\sqrt{2}}=5\sin\theta\), then \(\sin\theta=\frac{1}{\sqrt{2}}\), so \(\theta=\frac{\pi}{4}\).

Step4: Rewrite the integrand

Substitute \(x = 5\sin\theta\) into \(\frac{dx}{\sqrt{25 - x^{2}}}\), we get \(\frac{5\cos\theta d\theta}{\sqrt{25-25\sin^{2}\theta}}=\frac{5\cos\theta d\theta}{5\sqrt{1 - \sin^{2}\theta}}\). Since \(1-\sin^{2}\theta=\cos^{2}\theta\), the integrand becomes \(\frac{5\cos\theta d\theta}{5\cos\theta}=d\theta\).

Answer:

\(\int_{0}^{\frac{\pi}{4}}1\cdot d\theta\)