6int_0^pileft|left(sin3x+sin2x+sin xright)right|dx$6\int_{0}^{\pi}\left|\left(\sin3x+\sin2x+\sin x\right)\right|dx$ is equal to.....

### Related Formula sin A + sin B = 2 sinleft(fracA+B2right) cosleft(fracA-B2right)$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ sin 2x = 2 sin x cos x$$\sin 2x = 2 \sin x \cos x$$ cos 2x = 2 cos^2 x - 1$$\cos 2x = 2 \cos^2 x - 1$$ ### Core Logic Let I = 6int_0^pi|sin 3x + sin x + sin 2x| dx$I = 6\int_{0}^{\pi}|\sin 3x + \sin x + \sin 2x| dx$. Apply sum-to-product on sin 3x + sin x$\sin 3x + \sin x$: sin 3x + sin x = 2 sin(2x) cos(x)$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$ So the expression becomes: |2 sin(2x) cos x + sin 2x| = |sin 2x (2 cos x + 1)| = |2 sin x cos x (2 cos x + 1)|$|2 \sin(2x) \cos x + \sin 2x| = |\sin 2x (2 \cos x + 1)| = |2 \sin x \cos x (2 \cos x + 1)|$ Since x in [0, pi]$x \in [0, \pi]$, sin x geq 0$\sin x \geq 0$. We can pull it out of the modulus. I = 12 int_0^pi sin x |2 cos^2 x + cos x| dx$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$ ### Step 1: Coordinate Substitution Substitute t = cos x$t = \cos x$, then dt = -sin x dx$dt = -\sin x dx$. Limits: when x = 0$x = 0$, t = 1$t = 1$. When x = pi$x = \pi$, t = -1$t = -1$. I = 12 int_-1^1 |2t^2 + t| dt$$I = 12 \int_{-1}^{1} |2t^2 + t| dt$$ ### Step 2: Resolve Modulus Intervals The roots of 2t^2 + t = 0$2t^2 + t = 0$ are t = 0$t = 0$ and t = -1/2$t = -1/2$. The quadratic 2t^2 + t$2t^2 + t$ is negative in the interval (-1/2, 0)$(-1/2, 0)$ and positive elsewhere. Split the integral: I = 12 left[ int_-1^-1/2 (2t^2 + t) dt - int_-1/2^0 (2t^2 + t) dt + int_0^1 (2t^2 + t) dt right]$$I = 12 \left[ \int_{-1}^{-1/2} (2t^2 + t) dt - \int_{-1/2}^{0} (2t^2 + t) dt + \int_{0}^{1} (2t^2 + t) dt \right]$$

### Step 3: Evaluate Integrals Anti-derivative: F(t) = frac2t^33 + fract^22$F(t) = \frac{2t^3}{3} + \frac{t^2}{2}$. F(1) = 2/3 + 1/2 = 7/6$F(1) = 2/3 + 1/2 = 7/6$ F(0) = 0$F(0) = 0$ F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$ F(-1) = -2/3 + 1/2 = -1/6$F(-1) = -2/3 + 1/2 = -1/6$ Evaluate each segment: 1) int_-1^-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$\int_{-1}^{-1/2} = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$ 2) -int_-1/2^0 = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$-\int_{-1/2}^{0} = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$ 3) int_0^1 = F(1) - F(0) = 7/6 - 0 = 28/24$\int_{0}^{1} = F(1) - F(0) = 7/6 - 0 = 28/24$ Sum of parts inside bracket: frac524 + frac124 + frac2824 = frac3424 = frac1712$\frac{5}{24} + \frac{1}{24} + \frac{28}{24} = \frac{34}{24} = \frac{17}{12}$ ### Step 4: Final Output I = 12 times frac1712 = 17$$I = 12 \times \frac{17}{12} = 17$$ ### Pattern Recognition Whenever an integral features a cascading sum of sine frequencies like sin(kx)$\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = cos x$t = \cos x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions