Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
If $A = \int_0^{\sin \theta} \frac{t dt}{1 + t^2}$ and $B = \int_0^{\cos \theta} \frac{dt}{t(1 + t^2)}$, then the value of $e^A e^B \begin{vmatrix} A & A^2 & B \\ 1 & B^2 & -1 \\ 1 & A^2 + B^2 & -1 \end{vmatrix}$ is:
$\sin \theta$
$\csc \theta$
$0$
$1$
Step-by-Step Solution
Key Concept: Substitution $u = 1/x$ converts the integral into a form that reveals it equals the negative of another integral $B$.
For $A = \int_1^{\pi/2} \frac{t\,dt}{1+t^2}$, use substitution $u = \frac{1}{x}$ so $dt = -\frac{1}{x^2}dx$. The integral transforms to $-\int_1^{\cos 0} \frac{x^2 + 1}{x \cdot x^2}dx = -\int_1^{\cos 0} \frac{dx}{x(1+x^2)} = -B$. Thus $A = -B$.
Correct Answer: 3