How can you proof #intsqrt(x^2-a^2)dx# = #x/2sqrt(x^2-a^2)-a^2/2log|x+sqrt(x^2-a^2)|+C# using #x=asectheta#?

Question

#intsqrt(x^2-a^2)dx# = #x/2sqrt(x^2-a^2)-a^2/2log|x+sqrt(x^2-a^2)|# using #x=asectheta#

Charles Arocha · Accepted Answer

Let #x=a\sec	heta\implies dx=a\sec	heta	an	heta\ d	heta# #	herefore I=\int \sqrt{x^2-a^2}\ dx# #I=\int \sqrt{a^2\sec^2	heta-a^2}\ a\sec	heta	an	heta\ d	heta# #I=\int (a	an	heta)a\sec	heta	an	heta\ d	heta# #I=a^2\int \sec	heta	an^2	heta\ d	heta\ .........(1)# #I=a^2\int 	an	heta (\sec	heta	an	heta)\ d	heta# #I=a^2	an	heta\int \sec	heta	an	heta\ d	heta-a^2\int (d/{d	heta}	an	heta\cdot \int  \sec	heta	an	heta\ d	heta)d	heta# #I=a^2	an	heta\sec	heta-a^2\int sec^2	heta  \sec	heta\ d	heta# #I=a^2\sec	heta	an	heta-a^2\int (1+tan^2	heta)  \sec	heta\ d	heta# #I=a^2\sec	heta	an	heta-a^2\int  \sec	heta\ d	heta-a^2\int \sec	heta	an^2	heta\ d	heta# #I=a^2\sec	heta	an	heta-a^2\ln|\sec	heta+	an	heta|-I\ \quad (	ext{put I from(1)})# #2I=a^2\sec	heta	an	heta-a^2\ln|\sec	heta+	an	heta|+c# #I=1/2(a^2\sec	heta	an	heta-a^2\ln|\sec	heta+	an	heta|)+c_1# #I=1/2(a^2(x/a)\sqrt{x^2/a^2-1}-a^2\ln|x/a+\sqrt{x^2/a^2-1}|)+c_1# #I=1/2(x\sqrt{x^2-a^2}-a^2\ln|x+\sqrt{x^2-a^2}|+a^2\ln a)+c_1# #I=x/2 \sqrt{x^2-a^2}-{a^2}/2\ln|x+\sqrt{x^2-a^2}|+C# Proved.

Aurora Alvarado · Answer

undefined To prove the integral $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C$ using $x = a \sec \theta$, we will first express $dx$ in terms of $d\theta$ using the derivative of $x$ with respect to $\theta$.

Given $x = a \sec \theta$, we have $dx = a \sec \theta \tan \theta \, d\theta$.

Substitute $dx$ in the integral:

$$
\int \sqrt{(a \sec \theta)^2 - a^2} \cdot a \sec \theta \tan \theta \, d\theta
$$

Simplify under the square root:

$$
\int \sqrt{a^2 \sec^2 \theta - a^2} \cdot a \sec \theta \tan \theta \, d\theta
$$

$$
\int \sqrt{a^2 (\sec^2 \theta - 1)} \cdot a \sec \theta \tan \theta \, d\theta
$$

$$
\int \sqrt{a^2 \tan^2 \theta} \cdot a \sec \theta \tan \theta \, d\theta
$$

$$
\int a^2 \tan \theta \sec \theta \tan \theta \, d\theta
$$

$$
\int a^2 \tan^2 \theta \sec \theta \, d\theta
$$

Now, let's express the integral in terms of $\theta$ before integrating:

$$
a^2 \int \tan^2 \theta \sec \theta \, d\theta
$$

To simplify this integral, we'll use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$. Therefore:

$$
a^2 \int (\sec^2 \theta - 1) \sec \theta \, d\theta
$$

$$
a^2 \int \sec^3 \theta \, d\theta - a^2 \int \sec \theta \, d\theta
$$

Integrate $\int \sec^3 \theta \, d\theta$ using integration by parts, and integrate $\int \sec \theta \, d\theta$ directly. After integration, substitute back $x = a \sec \theta$ to arrive at the given result.