Find the indefinite integral.
${\int }_{}^{}\frac{se{c}^2\left(si{n}^{-1}\theta \right)}{\sqrt{1-{\theta }^2}}d\theta$

• A.
  $\tan \left(si{n}^{-1}\theta \right)+C$
• B. $\tan \left(co{s}^{-1}\theta \right)+C$
• C. $\cot \left(si{n}^{-1}\theta \right)+C$
• D. $si{n}^{-1}\left(tan\theta \right)+C$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize that the integral involves the composition of trigonometric and inverse trigonometric functions. Specifically, the term sec^2(sin^(-1)(θ)) suggests a relationship between the trigonometric and inverse trigonometric functions.
2. Step 2: Recall the derivative of tan(x), which is sec^2(x). This suggests that the integral may involve the tangent function. Additionally, the presence of sin^(-1)(θ) indicates that we need to use the substitution x = sin^(-1)(θ).
3. Step 3: Let x = sin^(-1)(θ). Then, θ = sin(x) and dθ = cos(x) dx. Also, note that √(1 - θ^2) = √(1 - sin^2(x)) = cos(x). Substitute these into the integral.
4. Step 4: After substitution, the integral becomes ∫sec^2(x) dx, because the √(1 - θ^2) in the denominator cancels with the cos(x) from dθ. The integral of sec^2(x) is tan(x).
5. Step 5: Substitute back x = sin^(-1)(θ) into the result to express the answer in terms of θ. The final result is tan(sin^(-1)(θ)) + C.