0606 P13 - Nov 2022 - Q7 - 5 marks
7878
Find the exact value of
\(\int_0^{\frac{\pi}{2}}\left(\cos3x+4\sin2x+1\right)\,dx.\)
Solution
Answer: \(\displaystyle \frac{11}{3}+\frac{\pi}{2}\).
Answer: \(\displaystyle \frac{11}{3}+\frac{\pi}{2}\).
Integrate term by term:
\(\int\left(\cos3x+4\sin2x+1\right)\,dx =\frac13\sin3x-2\cos2x+x.\)
Now apply the limits \(0\) and \(\frac{\pi}{2}\):
\(\left[\frac13\sin3x-2\cos2x+x\right]_0^{\frac{\pi}{2}}.\)
At \(x=\frac{\pi}{2}\),
\(\frac13\sin\frac{3\pi}{2}-2\cos\pi+\frac{\pi}{2} =-\frac13+2+\frac{\pi}{2} =\frac53+\frac{\pi}{2}.\)
At \(x=0\),
\(\frac13\sin0-2\cos0+0=-2.\)
Therefore the exact value of the integral is
\(\left(\frac53+\frac{\pi}{2}\right)-(-2) =\frac{11}{3}+\frac{\pi}{2}.\)
