Specialist2025 Exam 2 Section A · Q7CAS active1 mark

Integral by substitution u = cos θ

Using the substitution $u = \cos(\theta)$ , $\dfrac{1}{2}\displaystyle\int_0^{\frac{\pi}{2}} \frac{\sin(2\theta)}{1+\cos(\theta)}\,d\theta$ can be expressed as

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\displaystyle\int_0^{\frac{\pi}{2}} u\sqrt{\frac{1-u}{1+u}}\,du

\displaystyle\int_0^{1}\left(1+\frac{1}{1+u}\right)du

\displaystyle\int_0^{\frac{\pi}{2}}\left(1-\frac{1}{1+u}\right)du

\displaystyle\int_0^{1}\left(1-\frac{1}{1+u}\right)du

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