Question

If \[ \int \sin^2 t \tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx = A\sin^{-1}x + B\sqrt{1-x^2} + C, \] then \(A+B\) is equal to

• A. 10
• B. \(\frac{1}{2}\)
• C. 1
• D. \(-\frac{1}{2}\)

Hint:

Remember key identity: \(\tan^{-1}\sqrt{\frac{1-x}{1+x}} = \frac{1}{2}\cos^{-1}x\).

Answer 1

The Correct Option is C

Concept: Use substitution and standard identities.

Step 1: Use identity.
\[ \tan^{-1}\sqrt{\frac{1-x}{1+x}} = \frac{1}{2}\cos^{-1}x \]

Step 2: Simplify integral.
\[ \int \frac{1}{2}\cos^{-1}x \, dx \]

Step 3: Apply integration formula.
\[ \int \cos^{-1}x dx = x\cos^{-1}x - \sqrt{1-x^2} \]

Step 4: Compare with given form.
\[ A = 1,\quad B = 0 \Rightarrow A+B = 1 \]