A. \(\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]+K \quad\) B. \(\frac{1}{2} \theta-\frac{1}{2} \sin 2 \theta+k\) C. \(\frac{1}{2} \theta-\frac{1}{2} \cos 2 \theta+k\) D. \(\frac{1}{2} \theta-\frac{1}{4} \cos 2 \theta+k\)
Correct Answer: A
Explanation
can be written as: \(\int[\cos (90-\theta)]^{2}(90-\theta) d \theta\) Recall that \(\cos [90-\theta]=\sin \theta\). then. \(\int[\cos (90-\theta)]^{2} d \theta=\int(\sin \theta)^{2} d \theta\) recall that \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\) and that \(\sin ^{2} \theta+\cos ^{2} \theta=1\) from (ii), \(\cos ^{2} \theta=1-\sin ^{2} \theta\) put (iii) into (i) \(1-\sin ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\). \(\cos 2 \theta=1-2 \sin ^{2} \theta\) then, \(\sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\) then. \(\int \sin ^{2} \theta d \theta=\int \frac{1-\cos 2 \theta}{2}\) We then bring out the constant form out of the integral sign to have; \begin{array}{l} =\frac{1}{2}[1-\cos 2 \theta \\ =\frac{1}{2}\left[\theta-\frac{\cos 2 \theta}{2}\right]+k \\ =\frac{1}{2} \theta-\frac{\cos 2 \theta}{4}+k \end{array}