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

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

Answers (1)

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}

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