Question

Find the value of following integral $\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}$.
$\begin{aligned} & \text{a) }\dfrac{2}{3} \\\ & \text{b) }0 \\\ & \text{c) }\dfrac{-4}{3} \\\ & \text{d) }\dfrac{4}{3} \\\ \end{aligned}$

Answer 1

Now first we use the property $\left| x \right|=x$ if x > 0, and $\left| x \right|=-x$ if x < 0 and split the integral in two parts $\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\int\limits_{0}^{\dfrac{\pi}{2}}{{{\left| \cos x \right|}^{3}}dx}+\int\limits_{\dfrac{\pi }{2}}^{\pi }{{{\left| \cos x \right|}^{3}}dx}$. Now we can easily solve each integration as we know $\int{\cos x}=\sin x+C$.

Complete step-by-step solution:
First divide the integral in two parts as $\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left| \cos x \right|}^{3}}dx}+\int\limits_{\dfrac{\pi }{2}}^{\pi }{{{\left| \cos x \right|}^{3}}dx}$. We know that the function $\left| x \right|=x$ if x > 0, and $\left| x \right|=-x$ if x < 0.
Here for $0\le x\le \dfrac{\pi }{2},\cos x\ge 0$ and for $\dfrac{\pi }{2}\le x\le \pi ,\cos x\le 0$
Hence we get $\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( \cos x \right)}^{3}}dx}-\int\limits_{\dfrac{\pi }{2}}^{\pi }{{{\left( \cos x \right)}^{3}}dx}................(1)$
Now we know $\cos 3x=4{{\cos }^{3}}x-3\cos x$ . Rearranging the terms we get.
${{\cos }^{3}}x=\left( \dfrac{\cos 3x+3\cos x}{4} \right)$ .
Substituting this in equation (1) we get
$\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos 3x+3\cos x}{4} \right)dx}-\int\limits_{\dfrac{\pi }{2}}^{\pi }{\left( \dfrac{\cos 3x+3\cos x}{4} \right)dx}$
Now we know that $\int{cx=c\int{x}}$ hence taking the constant out of integration we get
$\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cos 3x+3\cos x \right)dx}-\dfrac{1}{4}\int\limits_{\dfrac{\pi }{2}}^{\pi }{\left( \cos 3x+3\cos x \right)dx}$
Now also we know that $\int{\left( f+g \right)}=\int{f}+\int{g}$ . Hence using this property we get
$\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cos 3x+3\cos x \right)dx}-\dfrac{1}{4}\int\limits_{\dfrac{\pi }{2}}^{\pi }{\left( \cos 3x+3\cos x \right)dx}$
$\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 3xdx}+\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{3\cos xdx}-\dfrac{1}{4}\int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos 3xdx}-\dfrac{1}{4}\int\limits_{\dfrac{\pi }{2}}^{\pi }{3\cos xdx}$
Now taking $\dfrac{1}{4}$ common from the equation we get
$\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\left[ \int\limits_{0}^{\dfrac{\pi }{2}}{\cos 3xdx}+\int\limits_{0}^{\dfrac{\pi }{2}}{3\cos xdx}-\int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos 3xdx}-\int\limits_{\dfrac{\pi }{2}}^{\pi }{3\cos xdx} \right]...........(2)$
Now let us first evaluate the integral $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 3x}dx$. Now we will use method of substitution to solve this method. Hence let us substitute $3x=t$ . Now differentiation on both sides we get $3dx=dt$ .
Hence we have $3x=t\Rightarrow dx=\dfrac{dt}{3}$ Also note that as $x\to 0,3x=t\to 0$ and as $x\to \dfrac{\pi }{2},3x=t\to \dfrac{3\pi }{2}$

& \int\limits_{0}^{\dfrac{\pi }{2}}{\cos 3x}=\int\limits_{0}^{\dfrac{3\pi }{2}}{\dfrac{\cos tdt}{3}} \\\ & =\dfrac{1}{3}{{\left[ \sin t \right]}^{\dfrac{3\pi }{2}}}_{0} \\\ & =\dfrac{1}{3}\left[ \sin \left( \dfrac{3\pi }{2} \right)-\sin 0 \right] \\\ & =\dfrac{1}{3}\left[ -1-0 \right] \\\ & =-\dfrac{1}{3} \\\ \end{aligned}$$ Hence we get $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 3xdx=-\dfrac{1}{3}}..................(3)$ Now let us evaluate $\int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos 3x}dx$. Now we will use method of substitution to solve this method. Hence let us substitute $3x=t$ . Now differentiation on both sides we get $3dx=dt$ . Hence we have $3x=t\Rightarrow dx=\dfrac{dt}{3}$ Also note that as $$x\to \pi ,3x=t\to 3\pi $$ and as $x\to \dfrac{\pi }{2},3x=t\to \dfrac{3\pi }{2}$ $$\begin{aligned} & \int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos 3x}=\int\limits_{\dfrac{3\pi }{2}}^{3\pi }{\dfrac{\cos tdt}{3}} \\\ & =\dfrac{1}{3}{{\left[ \sin t \right]}^{3\pi }}_{\dfrac{3\pi }{2}} \\\ & =\dfrac{1}{3}\left[ \sin \left( 3\pi \right)-\sin \left( \dfrac{3\pi }{2} \right) \right] \\\ & =\dfrac{1}{3}\left[ 0-(-1) \right] \\\ & =\dfrac{1}{3} \\\ \end{aligned}$$ Hence we get $\int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos 3xdx=\dfrac{1}{3}}..................(4)$ Now let us evaluate $\int\limits_{\dfrac{\pi }{2}}^{\pi }{3\cos x}dx$. $$\begin{aligned} & \int\limits_{\dfrac{\pi }{2}}^{\pi }{3\cos x}dx=3\int\limits_{\dfrac{\pi }{2}}^{\pi }{\cos x} \\\ & =3{{\left[ \sin x \right]}^{\pi }}_{\dfrac{\pi }{2}} \\\ & =3\left[ \sin \left( \pi \right)-\sin \left( \dfrac{\pi }{2} \right) \right] \\\ & =3\left[ 0-1 \right] \\\ & =-3 \\\ \end{aligned}$$ $\int\limits_{\dfrac{\pi }{2}}^{\pi }{3\cos xdx=-3}......................(5)$ Now let us evaluate $\int\limits_{0}^{\dfrac{\pi }{2}}{3\cos x}dx$. $$\begin{aligned} & \int\limits_{0}^{\dfrac{\pi }{2}}{3\cos x}dx=3\int\limits_{0}^{\dfrac{\pi }{2}}{\cos x} \\\ & =3{{\left[ \sin x \right]}^{\dfrac{\pi }{2}}}_{0} \\\ & =3\left[ \sin \left( \dfrac{\pi }{2} \right)-\sin \left( 0 \right) \right] \\\ & =3\left[ 1-0 \right] \\\ & =3 \\\ \end{aligned}$$ $$\int\limits_{0}^{\dfrac{\pi }{2}}{3\cos xdx=3}......................(6)$$ Now from equation (2), equation (3), equation (4), equation (5) and equation (6) we get. $$\begin{aligned} & \int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\left[ -\dfrac{1}{3}+3-\left( -3 \right)-\dfrac{1}{3} \right] \\\ & \int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\left[ 6-\dfrac{2}{3} \right] & \int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\left[ \dfrac{18-2}{3} \right] \\\ & \int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\dfrac{1}{4}\left[ \dfrac{16}{3} \right] \ & \int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}=\left[ \dfrac{4}{3} \right] \\\ \end{aligned}$$ Hence we get the value of $\int\limits_{0}^{\pi }{{{\left| \cos x \right|}^{3}}dx}$ = $\dfrac{4}{3}$ **Option d is the correct option.** **Note:** While using the method of substitution for integration note that the limits will also change. Hence if we substitute f(x) = t and the limits of x are a to b. then the limit of t is f(a) to f(b). Here we need to be careful while breaking the limit like for the mode function we have to find in which interval it is negative and positive.