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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 26 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 26 maths textbook solution.

Answers (1)

Answer:-  \frac{\pi }{4}-\frac{1}{2}

Hints:-  You must know the integral rules of trignometric  functions with its limits.

Given:-  \int_{-\pi / 4}^{\pi / 4} \sin ^{2} x \cdot d x

Solution : \int_{-\pi / 4}^{\pi / 4} \sin ^{2} x \cdot d x

            \begin{aligned} &=\int_{-\pi / 4}^{\pi / 4} \frac{1-\cos 2 x}{2} \cdot d x \\ &=\frac{1}{2} \int_{-\pi / 4}^{\pi / 4}(1-\cos 2 x) \cdot d x \\ &=\frac{1}{2}[x-\sin 2 x]_{-\pi / 4}^{\pi / 4} \end{aligned}

            \begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{4}-\sin 2 \times \frac{\pi}{4}-\left(\frac{-\pi}{4}\right)+\sin \left(2 \times \frac{-\pi}{4}\right)\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}-1+0\right] \\ &=\frac{2 \pi}{2 \times 4}-\frac{1}{2} \\ &=\frac{\pi}{4}-\frac{1}{2} \end{aligned}  

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