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

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}