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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.

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

Answers (1)

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

Now

                   \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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