#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 22 Maths Textbook Solution.

Answer: $\frac{3\pi }{16}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{4}xdx$

Solution:

$\int_{0}^{\frac{\pi }{2}}\cos ^{4}xdx=\int_{0}^{\frac{\pi }{2}}\left ( \cos ^{2}x \right )^{2}dx$

$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\cos 2 x}{2}\right)^{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} 2 \cos ^{2} x=1+\cos 2 x \\ \Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2} \end{array}\right]$

\begin{aligned} &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x)^{2} d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\cos ^{2} 2 x+2 \cos 2 x\right] d x \end{aligned}                                                        $\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\frac{1+\cos 4 x}{2}+2 \cos 2 x\right] d x \quad\left[\begin{array}{l} 2 \cos ^{2} x=1+\cos 2 x \\ \Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2} \end{array}\right]$

\begin{aligned} &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\frac{1}{2}+\frac{\cos 4 x}{2}+2 \cos 2 x\right] d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[\frac{3}{2}+\frac{\cos 4 x}{2}+2 \cos 2 x\right] d x \\ &=\frac{1}{4} \times \frac{3}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{1}{4 \times 2} \int_{0}^{\frac{\pi}{2}} \cos 4 x d x+\frac{2}{4} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \end{aligned}                                            $\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$

\begin{aligned} &=\frac{3}{8}[x]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{\sin 4 x}{4}\right]_{0}^{\frac{\pi}{2}}+\frac{2}{4}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{3}{8}\left[\frac{\pi}{2}-0\right]+\frac{1}{32}\left[\sin 4 \times \frac{\pi}{2}-\sin 4 \times 0\right]+\frac{1}{2 \times 2}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \\ &=\frac{3}{8}\left[\frac{\pi}{2}\right]+\frac{1}{32}[\sin 2 \pi-\sin 0]+\frac{1}{4}[\sin \pi-\sin 0] \\ &=\frac{3 \pi}{16}+0+0 \quad[\sin 2 \pi=\sin 0=\sin \pi=0] \\ &=\frac{3 \pi}{16} \end{aligned}