Trigonometry & Inverse Trigonometry
Definite Integral Involving cot inverse
nta_pyq_2024_apr
Grade 12
Question:
The integral $\displaystyle\int_{1/4}^{3/4}\cos\!\left(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\right)dx$ is equal to
$\dfrac{1}{2}$
$-\dfrac{1}{2}$
$-\dfrac{1}{4}$
$\dfrac{1}{4}$
Step-by-Step Solution
Key Concept: Use $\cot^{-1}\sqrt{\frac{1-x}{1+x}}=\tan^{-1}\sqrt{\frac{1+x}{1-x}}$. Then $\cos(2\tan^{-1}t)=\frac{1-t^2}{1+t^2}$ with $t=\sqrt{\frac{1+x}{1-x}}$. Simplify the expression.
Integrand simplifies to $-x$. $\int_{1/4}^{3/4}(-x)dx=-\frac{1}{4}$.
Correct Answer: 3