Integration Of Inverse Trigonometric Functions

Step 1: Simplify the integrand.
First, let's simplify the integrand $%5Cfrac%7Bx%20%2B%20%5Csqrt%7B1%20-%20x%5E2%7D%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D$ . We can break this up as: $%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%20%3D%20%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%20%2B%201$ Thus, the integral becomes: $%5Cint%20%5Csin%5E%7B-1%7Dx%20%5Cleft(%20%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%20%2B%201%20%5Cright)%20dx$

Step 2: Split the integral.
We can now split the integral into two parts: $%5Cint%20%5Csin%5E%7B-1%7Dx%20%5Ccdot%20%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%20-%20x%5E2%7D%7D%20%5C%2C%20dx%20%2B%20%5Cint%20%5Csin%5E%7B-1%7Dx%20%5C%2C%20dx$

Step 3: Solve the first integral.
We make the substitution $u%20%3D%20%5Csin%5E%7B-1%7Dx$ , so that $x%20%3D%20%5Csin%20u$ and $dx%20%3D%20%5Ccos%20u%20%5C%2C%20du$ . The first integral becomes: $%5Cint%20u%20%5Ccdot%20%5Cfrac%7B%5Csin%20u%7D%7B%5Ccos%20u%7D%20%5Ccdot%20%5Ccos%20u%20%5C%2C%20du%20%3D%20%5Cint%20u%20%5Csin%20u%20%5C%2C%20du$ This can be solved by integration by parts. Let $v%20%3D%20u$ and $dw%20%3D%20%5Csin%20u%20%5C%2C%20du$ , then: $%5Cint%20u%20%5Csin%20u%20%5C%2C%20du%20%3D%20-u%20%5Ccos%20u%20%2B%20%5Cint%20%5Ccos%20u%20%5C%2C%20du%20%3D%20-u%20%5Ccos%20u%20%2B%20%5Csin%20u$

Step 4: Solve the second integral.
The second integral is simply: $%5Cint%20%5Csin%5E%7B-1%7Dx%20%5C%2C%20dx%20%3D%20%5Cint%20u%20%5C%2C%20du%20%3D%20%5Cfrac%7Bu%5E2%7D%7B2%7D%20%2B%20C$

Step 5: Combine the results.
Thus, the original integral becomes: $%5Cboxed%7B-%5Csin%5E%7B-1%7Dx%20%5Ccos(%5Csin%5E%7B-1%7Dx)%20%2B%20%5Csin(%5Csin%5E%7B-1%7Dx)%20%2B%20%5Cfrac%7B(%5Csin%5E%7B-1%7Dx)%5E2%7D%7B2%7D%20%2B%20C%7D$

Concept: The principal value range of the inverse sine function is:

-%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Cle%20%5Csin%5E%7B-1%7D(x)%20%5Cle%20%5Cfrac%7B%5Cpi%7D%7B2%7D

We need the angle within this interval whose sine equals

-%5Cfrac%7B1%7D%7B2%7D

.
Step 1: Recall standard values

%5Csin%5Cleft(%5Cfrac%7B%5Cpi%7D%7B6%7D%5Cright)%20%3D%20%5Cfrac%7B1%7D%7B2%7D

Step 2: Use odd property of sine Since sine is an odd function:

%5Csin%5E%7B-1%7D(-x)%20%3D%20-%5Csin%5E%7B-1%7D(x)

So,

%5Csin%5E%7B-1%7D%5Cleft(-%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20-%5Csin%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D%5Cright)

Step 3: Substitute known value

%5Csin%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B6%7D

%5Ctherefore%20%5Csin%5E%7B-1%7D%5Cleft(-%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20-%5Cfrac%7B%5Cpi%7D%7B6%7D

Final Answer:

%5Cboxed%7B-%5Cfrac%7B%5Cpi%7D%7B6%7D%7D

Explanation: The angle must lie in the principal range

%5B-%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5D

, so the negative acute angle is chosen.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Integration Of Inverse Trigonometric Functions

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs