Definite Integrals Part 4

Concept:Use trigonometric identities to simplify the integrand, then break the integral at points where the expression changes sign.Explanation:Let $I = \int\limits_0^\pi |\sin 3x + \sin 2x + \sin x| \, dx$.Use identity: $\sin 3x + \sin x = 2 \sin 2x \cos x$.Thus, the integrand becomes $|\sin 2x (2\cos x + 1)|$.Find critical points in $[0,\pi]$ where either factor is zero.$\sin 2x = 0 \Rightarrow x = 0, \frac{\pi}{2}, \pi$$2\cos x + 1 = 0 \Rightarrow \cos x = -\frac12 \Rightarrow x = \frac{2\pi}{3}$Sign of $\sin 2x$: positive on $(0,\pi/2)$, negative on $(\pi/2,\pi)$.Sign of $2\cos x+1$: positive on $[0,2\pi/3)$, negative on $(2\pi/3,\pi]$.Therefore the product $(\sin 2x)(2\cos x+1)$ is:positive on $(0,\pi/2)$, negative on $(\pi/2,2\pi/3)$, positive on $(2\pi/3,\pi)$.Hence,$I = \int_0^{\pi/2} \sin 2x(2\cos x+1) dx - \int_{\pi/2}^{2\pi/3} \sin 2x(2\cos x+1) dx + \int_{2\pi/3}^{\pi} \sin 2x(2\cos x+1) dx$Rewrite $\sin 2x = 2\sin x \cos x$:$I = 2\int_0^{\pi/2} \sin x(2\cos^2 x+\cos x) dx - 2\int_{\pi/2}^{2\pi/3} \sin x(2\cos^2 x+\cos x) dx + 2\int_{2\pi/3}^{\pi} \sin x(2\cos^2 x+\cos x) dx$Substitute $t = \cos x$, $dt = -\sin x dx$.Limits: $x=0\to t=1$; $x=\pi/2\to t=0$; $x=2\pi/3\to t=-1/2$; $x=\pi\to t=-1$.This gives:$I = 2\int_1^0 (2t^2+t)(-dt) - 2\int_0^{-1/2} (2t^2+t)(-dt) + 2\int_{-1/2}^{-1} (2t^2+t)(-dt)$$= 2\int_0^1 (2t^2+t) dt - 2\int_{-1/2}^0 (2t^2+t) dt + 2\int_{-1}^{-1/2} (2t^2+t) dt$Integrate: $\int (2t^2+t) dt = \frac{2t^3}{3} + \frac{t^2}{2}$.Compute each part:First part: $\left[\frac{2t^3}{3}+\frac{t^2}{2}\right]_0^1 = \frac{2}{3}+\frac12 = \frac{7}{6} = \frac{28}{24}$Second part: $\left[\frac{2t^3}{3}+\frac{t^2}{2}\right]_{-1/2}^0 = 0 - \left(\frac{2}{3}(-\frac18)+\frac14\right) = -(-\frac{1}{12}+\frac14) = -\frac{1}{6} = -\frac{4}{24}$But careful: we have $I = 2[\text{first}] - 2[\text{second}] + 2[\text{third}]$. Actually we need to evaluate properly.Let's compute each integral with limits:$A = \int_0^1 (2t^2+t) dt = \left[\frac{2t^3}{3}+\frac{t^2}{2}\right]_0^1 = \frac{2}{3} + \frac12 = \frac{7}{6}$$B = \int_{-1/2}^0 (2t^2+t) dt = \left[\frac{2t^3}{3}+\frac{t^2}{2}\right]_{-1/2}^0 = 0 - \left(\frac{2}{3}(-\frac18) + \frac14\right) = -\left(-\frac{1}{12}+\frac14\right) = -\frac{1}{6}$$C = \int_{-1}^{-1/2} (2t^2+t) dt = \left[\frac{2t^3}{3}+\frac{t^2}{2}\right]_{-1}^{-1/2} = \left(\frac{2}{3}(-\frac18)+\frac14\right) - \left(\frac{2}{3}(-1) + \frac12\right) = \left(-\frac{1}{12}+\frac14\right) - \left(-\frac{2}{3}+\frac12\right) = \frac{1}{6} - \left(-\frac{1}{6}\right) = \frac{1}{3}$Thus $I = 2A - 2B + 2C = 2\left(\frac{7}{6}\right) - 2\left(-\frac{1}{6}\right) + 2\left(\frac{1}{3}\right) = \frac{7}{3} + \frac{1}{3} + \frac{2}{3} = \frac{10}{3}$? This seems inconsistent with the original solution. Let's re-examine the sign handling.In the original solution, they computed $I = 2\left[\left(\frac{2t^3}{3}+\frac{t^2}{2}\right)_{-1}^{-1/2} - \left(\frac{2t^3}{3}+\frac{t^2}{2}\right)_{-1/2}^0 + \left(\frac{2t^3}{3}+\frac{t^2}{2}\right)_0^1\right]$. That matches our expression because we had plus signs for the first and third, and minus for the second. Their bracket: first part (from -1 to -1/2), second part (from -1/2 to 0), third part (from 0 to 1). So the signs inside the bracket are + (first) - (second) + (third). Then multiplied by 2. Let's compute using their notation:Let $F(t)=\frac{2t^3}{3}+\frac{t^2}{2}$.First part: $F(-1/2)-F(-1)$Second part: $F(0)-F(-1/2)$Third part: $F(1)-F(0)$Then bracket = [F(-1/2)-F(-1)] - [F(0)-F(-1/2)] + [F(1)-F(0)] = F(-1/2)-F(-1)-F(0)+F(-1/2)+F(1)-F(0) = 2F(-1/2) - F(-1) + F(1) - 2F(0).Now compute:F(1)=2/3+1/2=7/6=28/24F(0)=0F(-1)=2/3(-1)+1/2(1)= -2/3+1/2 = -4/6+3/6=-1/6 = -4/24F(-1/2)=2/3(-1/8)+1/2(1/4)= -2/24+3/24=1/24So bracket = 2*(1/24) - (-1/6) + (7/6) - 0 = 2/24 + 1/6 + 7/6 = 1/12 + 8/6 = 1/12 + 16/12 = 17/12.Then I = 2 * (17/12) = 17/6.Hence 6I = 17.Thus the final answer is 17.Our earlier direct integral evaluation gave I=10/3? That's incorrect because we mis-signed the second integral sign. The correct expression from original is: $I = \int_0^{\pi/2} \sin2x(2\cos x+1)dx - \int_{\pi/2}^{2\pi/3} \sin2x(2\cos x+1)dx + \int_{2\pi/3}^{\pi} \sin2x(2\cos x+1)dx$. The substitution yields $I = 2\int_0^1 (2t^2+t) dt - 2\int_{-1/2}^0 (2t^2+t) dt + 2\int_{-1}^{-1/2} (2t^2+t) dt$. So we computed A=7/6, B=-1/6, C=1/3. Then I = 2A - 2B + 2C = 2*(7/6) - 2*(-1/6) + 2*(1/3) = 14/6 + 2/6 + 2/3 = 16/6 + 4/6 = 20/6 = 10/3. That gives 6I=20, not 17. So our direct evaluation must have a sign error. Let's re-check the substitution carefully.Original integrals:$I_1 = \int_0^{\pi/2} \sin2x(2\cos x+1) dx$$I_2 = \int_{\pi/2}^{2\pi/3} \sin2x(2\cos x+1) dx$$I_3 = \int_{2\pi/3}^{\pi} \sin2x(2\cos x+1) dx$Then $I = I_1 - I_2 + I_3$.Now substitute $t=\cos x$, $dt = -\sin x dx$, and $\sin2x = 2\sin x\cos x = 2t\sin x$. So integrand becomes $2t\sin x (2t+1) dx = 2t(2t+1) \sin x dx$. But $\sin x dx = -dt$. So the integrand becomes $2t(2t+1)(-dt) = -2t(2t+1) dt$.For $I_1$: x from 0 to π/2, t from 1 to 0. So $I_1 = \int_{t=1}^{0} -2t(2t+1) dt = \int_{0}^{1} 2t(2t+1) dt = 2\int_0^1 (2t^2+t) dt = 2A$.For $I_2$: x from π/2 to 2π/3, t from 0 to -1/2. So $I_2 = \int_{t=0}^{-1/2} -2t(2t+1) dt = -\int_{0}^{-1/2} 2t(2t+1) dt = \int_{-1/2}^{0} 2t(2t+1) dt = 2\int_{-1/2}^0 (2t^2+t) dt = 2B$ (where B is as before). But careful: B is integral from -1/2 to 0 of (2t^2+t) dt. So $I_2 = 2B$.For $I_3$: x from 2π/3 to π, t from -1/2 to -1. So $I_3 = \int_{t=-1/2}^{-1} -2t(2t+1) dt = -\int_{-1/2}^{-1} 2t(2t+1) dt = \int_{-1}^{-1/2} 2t(2t+1) dt = 2\int_{-1}^{-1/2} (2t^2+t) dt = 2C$ (where C is integral from -1 to -1/2). So $I_3 = 2C$.Then $I = I_1 - I_2 + I_3 = 2A - 2B + 2C$.Now compute A, B, C from earlier:$A = [\frac{2t^3}{3}+\frac{t^2}{2}]_0^1 = \frac{7}{6}$.$B = [\frac{2t^3}{3}+\frac{t^2}{2}]_{-1/2}^0 = 0 - (\frac{2}{3}(-\frac18)+\frac14) = -(-\frac{1}{12}+\frac14) = -(\frac{-1+3}{12}) = -\frac{2}{12} = -\frac{1}{6}$.Wait: $-\frac{1}{12}+\frac14 = -\frac{1}{12}+\frac{3}{12} = \frac{2}{12} = \frac{1}{6}$. So $F(-1/2) = -\frac{1}{12}+\frac14 = \frac{1}{6}$. Then B = 0 - 1/6 = -1/6. So indeed B = -1/6.$C = [\frac{2t^3}{3}+\frac{t^2}{2}]_{-1}^{-1/2} = F(-1/2) - F(-1)$. F(-1) = $\frac{2}{3}(-1)+\frac12(1) = -\frac{2}{3}+\frac12 = -\frac{4}{6}+\frac{3}{6} = -\frac{1}{6}$.So C = $\frac{1}{6} - (-\frac{1}{6}) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$.Now compute $I = 2A - 2B + 2C = 2(\frac{7}{6}) - 2(-\frac{1}{6}) + 2(\frac{1}{3}) = \frac{14}{6} + \frac{2}{6} + \frac{2}{3} = \frac{16}{6} + \frac{4}{6} = \frac{20}{6} = \frac{10}{3}$.Then $6I = 20$. But original says 17. So either the original solution has a mistake or we misinterpret the signs. Let's re-read the original solution's sign table: they said sign of (sin2x)(2cosx+1) is + on (0,π/2), - on (π/2,2π/3), + on (2π/3,π). So the absolute value converts negative to positive. That means $I = \int_0^{\pi/2} + \int_{\pi/2}^{2\pi/3} (-) + \int_{2\pi/3}^{\pi} +$, but careful: the absolute value makes the negative part positive by taking - of the integral. So $I = I_1 + (-)I_2 + I_3$? Actually if the function is negative on (π/2,2π/3), then $|f| = -f$, so integral over that interval is $-\int_{\pi/2}^{2\pi/3} f dx$. So $I = I_1 - I_2 + I_3$ where $I_1, I_2, I_3$ are integrals of f (without absolute) over the respective intervals. That's what we used.