#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 43

Answer:  $-\frac{1}{2 x}-\frac{1}{4} \sin \left(\frac{2}{x}\right)+c$

Hint: Use substitution method to solve this integral.

Given:   $\int \frac{1}{x^{2}} \cos ^{2}\left(\frac{1}{x}\right) d x$

Solution:

\begin{aligned} &\text { Let } I=\int \frac{1}{x^{2}} \cos ^{2}\left(\frac{1}{x}\right) d x \\ &\text { Put } \frac{1}{x}=t \Rightarrow-\frac{1}{x^{2}} d x=d t \\ &\Rightarrow d x=-x^{2} \text { dt then } \end{aligned}

$I=\int \frac{1}{x^{2}} \cdot \cos ^{2} t \cdot\left(-x^{2}\right) d t=-\int \cos ^{2} t \; d t$

$=-\int\left\{\frac{1+\cos 2 t}{2}\right\} d t$                                                    $\left[\begin{array}{l} \because 2 \cos ^{2} A-1=\cos 2 A \\ \Rightarrow 2 \cos ^{2} A=1+\cos 2 A \\ \Rightarrow \cos ^{2} A=\frac{1+\cos 2 A}{2} \end{array}\right]$

\begin{aligned} &=-\int\left\{\frac{1}{2}+\frac{\cos 2 t}{2}\right\} d t=-\int \frac{1}{2} d t-\frac{1}{2} \int \cos 2 t d t \\ &=-\frac{1}{2} \int 1 . d t-\frac{1}{2} \int \cos 2 t d t=-\frac{1}{2} \int t^{0} d t-\frac{1}{2} \int \cos 2 t d t \end{aligned}

$=-\frac{1}{2} t-\frac{1}{4} \sin 2\; t$                                                                $\left[\begin{array}{c} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \cos a\; x \; d x=\frac{\sin a x}{a}+c \end{array}\right]$

$=-\frac{1}{2 x}-\frac{1}{4} \sin \left(\frac{2}{x}\right)+c$