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  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 24 maths textbook solution.

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 24 maths textbook solution.

Answers (1)

Answer:- \frac{3\pi}{8}

Hints:-  You must know the integration rules of trigonometric functions and its limits.

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

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

    \begin{aligned} &=\int_{-\pi / 2}^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\cos ^{2} 2 x-2 \cos ^{2} 2 x\right) \cdot d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\frac{1+\cos ^{4} x}{2}-2 \cos 2 x\right) \cdot d x \end{aligned}

    \begin{aligned} &=\frac{3}{8} \int_{-\pi / 2}^{\pi / 2} d x+\frac{1}{8} \int \cos ^{4} x \cdot d x-\frac{1}{2} \int \cos 2 x \cdot d x \\ &\left.\left.\left.=\frac{3}{8} x .\right]_{-\pi / 2}^{\pi / 2}+\frac{1}{32} \sin ^{4} x\right]_{-\pi / 2}^{\pi / 2}-\frac{1}{2} \sin 2 x\right]_{-\pi / 2}^{\pi / 2} \\ &=\frac{3 \pi}{8} \end{aligned}

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