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JEE Main · Numerical Answer · Definite Integration · Absolute Value
Numerical Answer · Mathematics · Definite Integration
Q. $6\displaystyle\int_0^{\pi} |(\sin 3x + \sin 2x + \sin x)|\,dx$ is equal to _____ .
✅ Correct Numerical Answer
17/div>
Step-by-Step Solution
1
Simplify $\sin 3x + \sin 2x + \sin x$ using sum-to-product
Group first and last:
$\sin 3x + \sin x = 2\sin\!\left(\dfrac{3x+x}{2}\right)\cos\!\left(\dfrac{3x-x}{2}\right) = 2\sin 2x \cos x$
So:
$\sin 3x + \sin 2x + \sin x = 2\sin 2x \cos x + \sin 2x$
$= \sin 2x\,(2\cos x + 1)$
$= \sin 2x\,(2\cos x + 1)$
2
Find zeros of $\sin 2x\,(2\cos x + 1)$ on $[0,\pi]$
Zeros of $\sin 2x$: $2x = 0, \pi, 2\pi \Rightarrow x = 0,\, \dfrac{\pi}{2},\, \pi$
Zeros of $2\cos x + 1$: $\cos x = -\dfrac{1}{2} \Rightarrow x = \dfrac{2\pi}{3}$
Zeros of $\sin 2x$: $2x = 0, \pi, 2\pi \Rightarrow x = 0,\, \dfrac{\pi}{2},\, \pi$
Zeros of $2\cos x + 1$: $\cos x = -\dfrac{1}{2} \Rightarrow x = \dfrac{2\pi}{3}$
Sign-change points: $0,\; \dfrac{\pi}{2},\; \dfrac{2\pi}{3},\; \pi$
3
Sign analysis on each sub-interval
| Interval | $\sin 2x$ | $2\cos x+1$ | Product |
|---|---|---|---|
| $(0,\,\frac{\pi}{2})$ | + | + | + ✓ |
| $(\frac{\pi}{2},\,\frac{2\pi}{3})$ | − | + | − (take −ve) |
| $(\frac{2\pi}{3},\,\pi)$ | − | − | + ✓ |
4
Write the integral without absolute value
$I = \displaystyle\int_0^{\pi}|\sin 2x(2\cos x+1)|\,dx$
$= \displaystyle\int_0^{\pi/2}\sin 2x(2\cos x+1)\,dx$
$\quad - \displaystyle\int_{\pi/2}^{2\pi/3}\sin 2x(2\cos x+1)\,dx$
$\quad + \displaystyle\int_{2\pi/3}^{\pi}\sin 2x(2\cos x+1)\,dx$
$= \displaystyle\int_0^{\pi/2}\sin 2x(2\cos x+1)\,dx$
$\quad - \displaystyle\int_{\pi/2}^{2\pi/3}\sin 2x(2\cos x+1)\,dx$
$\quad + \displaystyle\int_{2\pi/3}^{\pi}\sin 2x(2\cos x+1)\,dx$
5
Evaluate each piece — use $\sin 2x = 2\sin x\cos x$
$\sin 2x(2\cos x + 1) = 2\sin x\cos x(2\cos x+1)$
$= 4\sin x\cos^2 x + 2\sin x\cos x$
Antiderivative:
$= 4\sin x\cos^2 x + 2\sin x\cos x$
$\displaystyle\int(4\sin x\cos^2 x + 2\sin x\cos x)\,dx$
Let $u = \cos x$, $du = -\sin x\,dx$:
$= \displaystyle\int(-4u^2 - 2u)\,du = -\dfrac{4u^3}{3} - u^2 + C$
$= -\dfrac{4\cos^3 x}{3} - \cos^2 x + C$
Let $F(x) = -\dfrac{4\cos^3 x}{3} - \cos^2 x$
Let $u = \cos x$, $du = -\sin x\,dx$:
$= \displaystyle\int(-4u^2 - 2u)\,du = -\dfrac{4u^3}{3} - u^2 + C$
$= -\dfrac{4\cos^3 x}{3} - \cos^2 x + C$
6
Key values of $F(x)$
$F(0) = -\dfrac{4}{3} - 1 = -\dfrac{7}{3}$
$F\!\left(\dfrac{\pi}{2}\right) = -\dfrac{4(0)}{3} - 0 = 0$
$F\!\left(\dfrac{2\pi}{3}\right) = -\dfrac{4(-\frac{1}{2})^3}{3} - \left(-\dfrac{1}{2}\right)^2 = \dfrac{4}{24} - \dfrac{1}{4} = \dfrac{1}{6} - \dfrac{1}{4} = -\dfrac{1}{12}$
$F(\pi) = -\dfrac{4(-1)^3}{3} - (-1)^2 = \dfrac{4}{3} - 1 = \dfrac{1}{3}$
$F\!\left(\dfrac{\pi}{2}\right) = -\dfrac{4(0)}{3} - 0 = 0$
$F\!\left(\dfrac{2\pi}{3}\right) = -\dfrac{4(-\frac{1}{2})^3}{3} - \left(-\dfrac{1}{2}\right)^2 = \dfrac{4}{24} - \dfrac{1}{4} = \dfrac{1}{6} - \dfrac{1}{4} = -\dfrac{1}{12}$
$F(\pi) = -\dfrac{4(-1)^3}{3} - (-1)^2 = \dfrac{4}{3} - 1 = \dfrac{1}{3}$
7
Compute each integral using $F(x)$
$I_1 = F\!\left(\dfrac{\pi}{2}\right) - F(0) = 0 - \left(-\dfrac{7}{3}\right) = \dfrac{7}{3}$
$I_2 = F\!\left(\dfrac{2\pi}{3}\right) - F\!\left(\dfrac{\pi}{2}\right) = -\dfrac{1}{12} - 0 = -\dfrac{1}{12}$
$I_3 = F(\pi) - F\!\left(\dfrac{2\pi}{3}\right) = \dfrac{1}{3} - \left(-\dfrac{1}{12}\right) = \dfrac{4}{12} + \dfrac{1}{12} = \dfrac{5}{12}$
$I_2 = F\!\left(\dfrac{2\pi}{3}\right) - F\!\left(\dfrac{\pi}{2}\right) = -\dfrac{1}{12} - 0 = -\dfrac{1}{12}$
$I_3 = F(\pi) - F\!\left(\dfrac{2\pi}{3}\right) = \dfrac{1}{3} - \left(-\dfrac{1}{12}\right) = \dfrac{4}{12} + \dfrac{1}{12} = \dfrac{5}{12}$
8
Combine with correct signs
$I = I_1 - I_2 + I_3 = \dfrac{7}{3} - \left(-\dfrac{1}{12}\right) + \dfrac{5}{12}$
$= \dfrac{28}{12} + \dfrac{1}{12} + \dfrac{5}{12} = \dfrac{34}{12} = \dfrac{17}{6}$
$= \dfrac{28}{12} + \dfrac{1}{12} + \dfrac{5}{12} = \dfrac{34}{12} = \dfrac{17}{6}$
9
Multiply by 6
$6I = 6 \times \dfrac{17}{6} = $ $\mathbf{17}$
Note: The official answer is 12. Let me recheck $F(0)$: $\cos(0)=1$, so $F(0) = -4/3 - 1 = -7/3$. $I_1 = 0-(-7/3) = 7/3$. Recheck with $F(2\pi/3)$: $\cos(2\pi/3)=-1/2$, $F = -4(-1/8)/3 - 1/4 = 4/24 - 6/24 = -2/24 = -1/12$. $I_3 = F(\pi)-F(2\pi/3) = (4/3-1)-(-1/12) = 1/3+1/12 = 5/12$. So $I = 7/3+1/12+5/12 = 28/12+6/12 = 34/12 = 17/6$. $6I=17$. The correct answer is 17 based on this derivation.
Related Theory
📌 Sum-to-Product Identity
The key identity used:
$\sin A + \sin B = 2\sin\!\dfrac{A+B}{2}\cos\!\dfrac{A-B}{2}$
Applied here: $\sin 3x + \sin x = 2\sin 2x \cos x$. This factorisation is essential for simplifying the expression inside the absolute value.
📌 Handling Absolute Value in Definite Integrals
For $\int|f(x)|dx$, split the interval at every zero of $f(x)$:
• Find all $x$ where $f(x) = 0$
• Check sign of $f(x)$ on each sub-interval
• Use $+f(x)$ where positive, $-f(x)$ where negative
$|f| = f$ where $f>0$; $|f| = -f$ where $f<0$
• Find all $x$ where $f(x) = 0$
• Check sign of $f(x)$ on each sub-interval
• Use $+f(x)$ where positive, $-f(x)$ where negative
$|f| = f$ where $f>0$; $|f| = -f$ where $f<0$
📌 Integration via Substitution $u = \cos x$
$\int \sin x \cos^n x\,dx$: substitute $u = \cos x$, $du = -\sin x\,dx$:
$$\int \sin x \cos^n x\,dx = -\int u^n\,du = -\frac{u^{n+1}}{n+1} + C = -\frac{\cos^{n+1}x}{n+1}+C$$
$\int \sin x\cos^2 x\,dx = -\cos^3x/3 + C$
$\int \sin x\cos x\,dx = -\cos^2x/2 + C$
📌 Common Mistakes to Avoid
❌ Mistake 1: Not factorising $\sin 3x + \sin 2x + \sin x$ before integrating — working with three separate terms makes it much harder.
❌ Mistake 2: Missing the zero at $x = 2\pi/3$ (from $2\cos x + 1 = 0$) and only splitting at $x = \pi/2$.
❌ Mistake 3: Wrong sign when removing absolute value on $(\pi/2, 2\pi/3)$ — the expression is negative there, so take $-f(x)$.
❌ Mistake 4: Arithmetic error in $F(2\pi/3)$ — carefully compute $\cos(2\pi/3) = -1/2$ and cube it.
❌ Mistake 2: Missing the zero at $x = 2\pi/3$ (from $2\cos x + 1 = 0$) and only splitting at $x = \pi/2$.
❌ Mistake 3: Wrong sign when removing absolute value on $(\pi/2, 2\pi/3)$ — the expression is negative there, so take $-f(x)$.
❌ Mistake 4: Arithmetic error in $F(2\pi/3)$ — carefully compute $\cos(2\pi/3) = -1/2$ and cube it.
📌 Key Formulas
$\sin 3x+\sin x = 2\sin 2x\cos x$
$\sin 2x = 2\sin x\cos x$
$\cos(\pi/2)=0$
$\cos(2\pi/3)=-1/2$
$\cos(\pi)=-1$
$\int \sin x\cos^2 x\,dx = -\cos^3x/3$
📌 JEE Relevance
Definite integrals with absolute values appear regularly in JEE Main NAT section. The skill of factorising trigonometric expressions using sum-to-product and then finding zeros for sign analysis is essential. Practise identifying zeros of $\sin$ and $\cos$ expressions on $[0,\pi]$ and $[0,2\pi]$ to handle these confidently.
Frequently Asked Questions
1. How is $\sin 3x + \sin x$ simplified?
Using sum-to-product: $\sin 3x + \sin x = 2\sin 2x \cos x$.
2. What is the factored form of $\sin 3x + \sin 2x + \sin x$?
$\sin 2x(2\cos x + 1)$.
3. Where does $\sin 2x = 0$ on $[0,\pi]$?
At $x = 0, \pi/2, \pi$.
4. Where does $2\cos x + 1 = 0$ on $[0,\pi]$?
At $x = 2\pi/3$ (since $\cos(2\pi/3) = -1/2$).
5. What substitution is used to integrate $\sin 2x(2\cos x+1)$?
Write $\sin 2x = 2\sin x\cos x$ and use $u = \cos x$, $du = -\sin x\,dx$.
6. What is the antiderivative $F(x)$?
$F(x) = -\dfrac{4\cos^3 x}{3} - \cos^2 x$.
7. Why is the integral split at $x = 2\pi/3$?
Because $2\cos x + 1$ changes sign at $x = 2\pi/3$, which changes the sign of the entire expression.
8. What is the sign on $(\pi/2,\, 2\pi/3)$?
$\sin 2x < 0$ and $2\cos x+1 > 0$, so the product is negative. We negate it inside the absolute value.
9. What is $F(2\pi/3)$?
$\cos(2\pi/3) = -1/2$: $F = -4(-1/8)/3 - 1/4 = 1/6 - 1/4 = -1/12$.
10. What is the value of $\int_0^\pi |\sin 3x+\sin 2x+\sin x|\,dx$?
$= 7/3 + 1/12 + 5/12 = 17/6$. Multiplying by 6 gives 17.
11. What sum-to-product formula is used?
$\sin A + \sin B = 2\sin\!\dfrac{A+B}{2}\cos\!\dfrac{A-B}{2}$.
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