Find the indefinite integral.∫sec2(sin−1θ)1−θ2dθ\int_{}^{}\frac{sec^2(sin^{-1}\theta)}{\sqrt{1-\theta^2}}d\theta

tan ⁡ ( s i n − 1 θ ) + C \tan\left(sin^{-1}\theta\right)+C

Find the indefinite integral. ∫ s e c 2 ( s i n − 1 θ ) 1 − θ 2 d θ \int_{}^{}\frac{sec^2(sin^{-1}\theta)}{\sqrt{1-\theta^2}}d\theta

we're asked to find the indefinite integral of the secant squared of the inverse sine of theta over the square root of one minus theta squared d theta. Now to solve this integral, I can tell that I need to do a substitution, but the question is what should I choose u to be when making that substitution? Now looking at this integral, I see that I have this inverse sine and I also have the square root of one minus theta squared in my denominator here, which results when taking the derivative of the inverse sine. So that's a good clue that I should choose u as the inverse sign of theta. Now if u is equal to the inverse sign of theta, that makes d u equal to one over the square root of one minus theta squared d theta. So in this integral, having chosen this as u, I then have du already appear there. Now I can rewrite this as the integral of the secant squared of u du having made that substitution. So this is then going to be equal to the tangent of u plus c, and I can plug u back in here. I know it's the inverse sine. So this is the tangent of the inverse sine of theta and then plus c, and this is my final answer here. Let us know if you have any questions, and I'll see you in the next one.

Find the indefinite integral. ∫ s e c 2 ⁡ ( s i n − 1 θ ) 1 − θ 2 d θ \int_{}^{}\frac{sec^2⁡(sin^{-1}\theta)}{\sqrt{1-\theta^2}}d\theta

tan ⁡ ( s i n − 1 θ ) + C \tan\left(sin^{-1}\theta\right)+C

tan ⁡ ( c o s − 1 θ ) + C \tan\left(cos^{-1}\theta\right)+C

cot ⁡ ( s i n − 1 θ ) + C \cot\left(sin^{-1}\theta\right)+C

s i n − 1 ( t a n θ ) + C sin^{-1}(tan\theta)+C

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the indefinite integral.
$\int_{}^{} \frac{s e c^{2} (s i n^{- 1} θ)}{\sqrt{1 - θ^{2}}} d θ$

$\tan (s i n^{- 1} θ) + C$

$\tan (c o s^{- 1} θ) + C$

$\cot (s i n^{- 1} θ) + C$

$s i n^{- 1} (t a n θ) + C$

Verified step by step guidance

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.

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).

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.

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).

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.

4:51m

Watch next

Master Integrals Resulting in Inverse Trig Functions with a bite sized video explanation from Patrick

Start learning

Integrals Involving Inverse Trigonometric Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice