Specialist2025 NHT Exam 2 Section A · Q8CAS active1 mark

Trigonometric substitution in a definite integral

Using the substitution $x = \tan(\theta)$ , then $\displaystyle\int_{-1}^{1} \frac{1}{(1+x^2)^2}\,dx$ is equivalent to

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\displaystyle\int_{0}^{\frac{\pi}{4}} \frac{1}{\sec^2(\theta)}\,d\theta

\displaystyle\int_{0}^{\frac{\pi}{4}} 2\cos^2(\theta)\,d\theta

\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 2\cos^4(\theta)\,d\theta

\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\sec^4(\theta)}\,d\theta

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