Step 1: Identify the form of the integral.
This integral is of the form ∫x2−a2dx, which suggests using a trigonometric substitution.
Step 2: Choose the appropriate trigonometric substitution.
For expressions involving x2−a2, the standard substitution is x=asecθ.
From this substitution, we can find dx:
dx=asecθtanθdθ
Also, we can express x2−a2 in terms of θ:
x2−a2=(asecθ)2−a2=a2sec2θ−a2=a2(sec2θ−1)=a2tan2θ=a∣tanθ∣
Assuming x>a (or x<−a) and θ is in an appropriate range where tanθ≥0, we have x2−a2=atanθ.
Step 3: Substitute into the integral.
∫(atanθ)(asecθtanθ)dθ∫a2tan2θsecθdθ
Step 4: Rewrite tan2θ using the identity tan2θ=sec2θ−1.
∫a2(sec2θ−1)secθdθa2∫(sec3θ−secθ)dθa2(∫sec3θdθ−∫secθdθ)
Step 5: Evaluate the standard integrals.
We use the known integral formulas:
∫secθdθ=ln∣secθ+tanθ∣+C1∫sec3θdθ=21secθtanθ+21ln∣secθ+tanθ∣+C2
Substitute these back into the expression:
a2((21secθtanθ+21ln∣secθ+tanθ∣)−ln∣secθ+tanθ∣)+Ca2(21secθtanθ−21ln∣secθ+tanθ∣)+C2a2(secθtanθ−ln∣secθ+tanθ∣)+C
Step 6: Convert back to x.
From x=asecθ, we have secθ=ax.
To find tanθ, we use the identity tanθ=sec2θ−1:
tanθ=(ax)2−1=a2x2−1=a2x2−a2=ax2−a2
Now substitute these expressions back into the result:
2a2(ax⋅ax2−a2−lnax+ax2−a2)+C2a2(a2xx2−a2−lnax+x2−a2)+C
Distribute 2a2:
2xx2−a2−2a2lnax+x2−a2+C
Using the logarithm property lnBA=ln∣A∣−ln∣B∣:
2xx2−a2−2a2(ln∣x+x2−a2∣−ln∣a∣)+C2xx2−a2−2a2ln∣x+x2−a2∣+2a2ln∣a∣+C
Since 2a2ln∣a∣ is a constant, it can be absorbed into the arbitrary constant C.
The final answer is 2xx2−a2−2a2ln∣x+x2−a2∣+C.
That's 2 down. 3 left today — send the next one.
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This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
To solve the integral sqrt(x^2 - a^2) \, dx: Step 1: Identify the form of the integral. This integral is of the form sqrt(x^2 - a^2) \, dx, which suggests using a trigonometric substitution. Step 2: Choose the appropriate trigonometric substitution. For expressions involving sqrt(x^2 - a^2), the standard substitution is x = a . From this substitution, we can find dx: dx = a \, d Also, we can express sqrt(x^2 - a^2) in terms of : sqrt(x^2 - a^2) = sqrt((a )^2 - a^2) = sqrt(a^2 ^2 - a^2) = sqrt(a^2 (^2 - 1)) = sqrt(a^2 ^2 ) = a | | Assuming x > a (or x < -a) and is in an appropriate range where 0, we have sqrt(x^2 - a^2) = a . Step 3: Substitute into the integral. (a ) (a ) \, d a^2 ^2 \, d Step 4: Rewrite ^2 using the identity ^2 = ^2 - 1. a^2 (^2 - 1) \, d a^2 (^3 - ) \, d a^2 ( ^3 \, d - \, d ) Step 5: Evaluate the standard integrals. We use the known integral formulas: \, d = | + | + C_1 ^3 \, d = (1)/(2) + (1)/(2) | + | + C_2 Substitute these back into the expression: a^2 ( ( (1)/(2) + (1)/(2) | + | ) - | + | ) + C a^2 ( (1)/(2) - (1)/(2) | + | ) + C (a^2)/(2) ( - | + |) + C Step 6: Convert back to x. From x = a , we have = (x)/(a). To find , we use the identity = sqrt(^2 - 1): = sqrt(((x)/(a))^2 - 1) = sqrt((x^2)/(a^2) - 1) = sqrt((x^2 - a^2)/(a^2)) = sqrt(x^2 - a^2)a Now substitute these expressions back into the result: (a^2)/(2) ( (x)/(a) · sqrt(x^2 - a^2)a - |(x)/(a) + sqrt(x^2 - a^2)a| ) + C (a^2)/(2) ( x sqrt(x^2 - a^2)a^2 - |x + sqrt(x^2 - a^2)a| ) + C Distribute (a^2)/(2): x sqrt(x^2 - a^2)2 - (a^2)/(2) |x + sqrt(x^2 - a^2)a| + C Using the logarithm property |(A)/(B)| = |A| - |B|: x sqrt(x^2 - a^2)2 - (a^2)/(2) (|x + sqrt(x^2 - a^2)| - |a|) + C x sqrt(x^2 - a^2)2 - (a^2)/(2) |x + sqrt(x^2 - a^2)| + (a^2)/(2) |a| + C Since (a^2)/(2) |a| is a constant, it can be absorbed into the arbitrary constant C. The final answer is (x)/(2)sqrt(x^2 - a^2) - (a^2)/(2)|x + sqrt(x^2 - a^2)| + C. That's 2 down. 3 left today — send the next one.
