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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 16 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 16 Maths Textbook Solution.

Answers (1)

Answer: \frac{x^{3}}{6}-\frac{x^{2} \sin 2 x}{4}-\frac{x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c

Hint: Use the formula

          1) \sin ^{2} x=\frac{1-\cos 2 x}{2}

          \text { 2) } \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x

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

Solution: I=\int x^{2} \sin ^{2} x d x

              \begin{aligned} &\int x^{2}\left(\frac{1-\cos 2 x}{2}\right) d x=\int \frac{x^{2}}{2}-\frac{x^{2} \cos 2 x}{2} d x \\ &=\int \frac{x^{2}}{2} d x-\int \frac{x^{2} \cos 2 x}{2} d x \\ &=\frac{x^{3}}{3 \times 2}-\frac{1}{2} \int x^{2} \cos 2 x d x \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \int \cos 2 x d x-\int\left(\frac{d}{d x} x^{2} \int \cos 2 x d x\right) d x\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\int 2 x \frac{\sin 2 x}{2} d x\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\int x \sin 2 x d x\right) \end{aligned}

Now by integrating the second part we have,

               \begin{aligned} &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}-\left[x \frac{-\cos 2 x}{2}-\int 1 \frac{-\cos 2 x}{2} d x\right]\right) \\ &=\frac{x^{3}}{6}-\frac{1}{2}\left(x^{2} \frac{\sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{\sin 2 x}{4}\right)+c \\ &=\frac{x^{3}}{6}-\frac{x^{2} \sin 2 x}{4}-\frac{x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c \end{aligned}

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