6 _ ^0|( 3x+ 2x+ x)|dx is equal to ____ .

Question

$6{\int }_{\pi }^{0}|(\sin 3x+\sin 2x+\sin x)|dx$ is equal to $____$ .

Answer: 17

Accepted Answer

Let, $I=\int _{0}^{\pi }|\sin 3x+\sin 2x+\sin x|dx$

We know, $\sin 3x+\sin x=2\sin 2x\cos x$

$	herefore $ $\sin 3x+\sin 2x+\sin x=2\sin 2x\cos x+\sin 2x=\sin 2x(2\cos x+1)$

$I=\int _{0}^{\pi }|\sin 2x(2\cos x+1)|dx$

Critical points : $\forallx\in [0,\pi ]$

$\sin 2x=0\Rightarrow x=0,\frac{\pi }{2},\pi $

$\Rightarrow$ $2\cos x+1=0\Rightarrow \cos x=-\frac{1}{2}\Rightarrow x=\frac{2\pi }{3}$

$\begin{gathered} 2\cos x+1>0 \forallx\in [0,2\pi /3) \ 2\cos x+1\le 0 \forallx\in [2\pi /3,\pi ] \end{gathered}$

Interval

sign of sin 2x

sign of 2 cos x + 1

sign of (sin 2x)(2 cos x + 1)

(0, \pi /2)

+

(\pi /2, 2\pi /3)

-

+

-

(2\pi /3, \pi )

-

+

$I=\int _{0}^{\pi /2}\sin 2x(2\cos x+1)dx-\int _{\pi /2}^{2\pi /3}\sin 2x(2\cos x+1)dx+\int _{2\pi /3}^{\pi }\sin 2x(2\cos x+1)dx$

Using, $\sin 2x=2\sin x\cos x$

$I=\int _{0}^{\pi /2}2\sin x(2{\cos }^{2}x+\cos x)dx-\int _{\pi /2}^{2\pi /3}2\sin x(2{\cos }^{2}x+\cos x)dx+\int _{2\pi /3}^{\pi }2\sin x(2{\cos }^{2}x+\cos x)dx$

Let $t=\cos x,\sin xdx=-dt$, limits get change

$I=\int _{1}^{0}-2(2{t}^{2}+t)dt+\int _{0}^{-1/2}2(2{t}^{2}+t)dt-\int _{-1/2}^{-1}2(2{t}^{2}+t)dt$

$\Rightarrow$ $I=2\int _{0}^{1}(2{t}^{2}+t)dt-2\int _{-1/2}^{0}(2{t}^{2}+t)dt+2\int _{-1}^{-1/2}(2{t}^{2}+t)dt$

$\Rightarrow$ $I=2[{(\frac{2{t}^{3}}{3}+\frac{{t}^{2}}{2})}_{-1/2}^{-1}-{(\frac{2{t}^{3}}{3}+\frac{{t}^{2}}{2})}_{0}^{-1/2}+{(\frac{2{t}^{3}}{3}+\frac{{t}^{2}}{2})}_{1}^{0}]$

$\begin{gathered} =[\frac{2}{3}(-1/2{)}^{3}+\frac{1}{2}(-1/2{)}^{2}-{\frac{2}{3}(-1{)}^{3}+\frac{1}{2}(-1{)}^{2}}]=[-1/12+1/8+2/3-1/2]=\frac{5}{24} \ =[\frac{2}{3}(0)+0-{\frac{2}{3}(-1/2{)}^{3}+\frac{1}{2}(-1/2{)}^{2}}]=[1/12-1/8]=-\frac{1}{24} \ =[\frac{2}{3}(1{)}^{3}+\frac{1}{2}(1{)}^{2}-{0+0}]=\frac{7}{6}=\frac{28}{24} \end{gathered}$

$\Rightarrow$ $I=2[\frac{5}{24}-(-1/24)+\frac{28}{24}]=2	imes \frac{34}{24}=\frac{17}{6}$

$\Rightarrow 6I=17$

Solutions · 1