rewrite the given integral using this substitution.\n \frac{5}{sqrt{2}}int_{0}^{}\frac{dx}{sqrt{25…

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)