Question

The value of the integral ${\int }_{\pi }^{2\pi }\frac{(x^2+2)\cos x}{x^3}dx$ is

• A. $-\left(\frac{5}{4{\pi }^2}\right)$
• B. $-\left(\frac{1}{4{\pi }^2}\right)$
• C. $\frac{1}{4{\pi }^2}$
• D. $\frac{5}{4{\pi }^2}$

Answer 1

Answer: (A)

$-\left(\frac{5}{4{\pi }^2}\right)$

Given Integral, $I={\int }_{\pi }^{2\pi }\frac{(x^2+2)cosx}{x^3}dx$

$\to I={\int }_{\pi }^{2\pi }\frac{cosx}{x}dx+{\int }_{\pi }^{2\pi }\frac{2cosx}{x^3}dx$

$\to {\int }_{\pi }^{2\pi }\frac{2cosx}{x^3}dx=$ ${\int }_{\pi }^{2\pi }\underset{I}{\underset{}{\left(2cosx\right)}}\underset{II}{\underset{}{\left(\frac{1}{x^3}\right)}}dx$

We know that $\int f_I\left(x\right)f_{II}\left(x\right)dx=f_I\left(x\right).\int f{(}_{II}x)dx-\int f_I^{\prime }\left(x\right)\left(\int f_{II}\left(x\right)dx\right)dx$

From eq. $\left(1\right)$ we can understand $f_I\left(x\right)=2cosx$ and $f_{II}\left(x\right)=\frac{1}{x^3}$

$\to$ ${\int }_{\pi }^{2\pi }\underset{I}{\underset{}{\left(2cosx\right)}}\underset{II}{\underset{}{\left(\frac{1}{x^3}\right)}}dx=$ $2cosx.{\int }_{\pi }^{2\pi }\left(\frac{1}{x^3}\right)dx-{\int }_{\pi }^{2\pi }(-2sinx)\left({\int }_{\pi }^{2\pi }\frac{1}{x^3}dx\right)dx$

$\to {\int }_{\pi }^{2\pi }\frac{2cosx}{x^3}dx=$ $2cosx(-\frac{1}{{2x}^2})-{\int }_{\pi }^{2\pi }(-2sinx)(-\frac{1}{{2x}^2})dx$

$\Rightarrow (-\frac{cosx}{x^2})-{\int }_{\pi }^{2\pi }\left(\frac{sinx}{x^2}\right)dx$

$\Rightarrow (-\frac{cosx}{x^2})-{\int }_{\pi }^{2\pi }sinx.\left(\frac{1}{x^2}\right)dx$

Using the integration by parts formula again, $\int f_I\left(x\right)f_{II}\left(x\right)dx=f_I\left(x\right).{\int }_{\pi }^{2\pi }f{(}_{II}x)dx-\int f_I^{\prime }\left(x\right)\left({\int }_{\pi }^{2\pi }{f}_{II}\left(x\right)dx\right)dx$

$\Rightarrow (-\frac{cosx}{x^2})-(sinx.\left(\frac{-1}{x}\right)-{\int }_{\pi }^{2\pi }cosx.\left(\frac{-1}{x}\right))dx$

$\Rightarrow (-\frac{cosx}{x^2})+\left(\frac{sinx}{x}\right)-{\int }_{\pi }^{2\pi }\left(\frac{cosx}{x}\right)dx$

Hence, ${\int }_{\pi }^{2\pi }\frac{2cosx}{x^3}dx=$$(-\frac{cosx}{x^2})+\left(\frac{sinx}{x}\right)-{\int }_{\pi }^{2\pi }\left(\frac{cosx}{x}\right)dx$

We know that $\to I={\int }_{\pi }^{2\pi }\frac{cosx}{x}dx+{\int }_{\pi }^{2\pi }\frac{2cosx}{x^3}dx$

So now $\to I={\int }_{\pi }^{2\pi }\frac{cosx}{x}dx+$ $(-\frac{cosx}{x^2})+\left(\frac{sinx}{x}\right)-\int \left(\frac{cosx}{x}\right)dx$

$I={(-\frac{cosx}{x^2})}_{\pi }^{2\pi }+{\left(\frac{sinx}{x}\right)}_{\pi }^{2\pi }$

$I=-\frac{1}{4{\pi }^2}-\frac{1}{{\pi }^2}+(0-0)$

$I=-\left(\frac{5}{4{\pi }^2}\right)$

Hence correct answer is $A$.