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

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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


        \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


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