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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 13 Maths Textbook Solution.

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Answer: \frac{x^{2}}{4}+\frac{x \sin (2 x)}{4}+\frac{\cos (2 x)}{8}+c

Hint:Using integration by parts

         \int a b d x=a \int b d x-\int\left[\frac{d}{d x} a \int b d x\right] d x

Given: I=\int x \cos ^{2} x d x

Solution: \int x \cos ^{2} x d x=\int x \cdot \frac{1+\cos 2 x}{2} d x

 Break this integral apart now

                            \begin{aligned} &=\int x \cdot\left(\frac{1+\cos 2 x}{2}\right) d x \\ &=\int \frac{x}{2} d x+\int \frac{x \cos (2 x)}{2} d x \end{aligned}

Evaluate the integral part by part

                     \begin{aligned} &\int \frac{x}{2} d x=\frac{x^{2}}{4}+c \\ &\int \frac{x \cos (2 x)}{2} d x \end{aligned}

Integration by parts

       Let u=\frac{\cos (2 x)}{2}

             \int u d x=\frac{\sin 2 x}{4}

\int \frac{x \cos (2 x)}{2} d x=\frac{x \cdot \sin (2 x)}{4}-\int \frac{\sin 2 x}{4} d x


                   \int \frac{-\sin (2 x)}{4} d x=\frac{\cos (2 x)}{8}+c

Combining all,

           \int x \cos ^{2}(x) d x=\frac{x^{2}}{4}+\frac{x \cdot \sin (2 x)}{4}+\frac{\cos (2 x)}{8}+c

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