#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 23 Maths Textbook Solution.

Answer: $\frac{\pi }{4}\left ( a^{2}+b^{2} \right )$

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

Given:$\int_{0}^{\frac{\pi }{2}}\left ( a^{2}\cos ^{2}x+b^{2}\sin ^{2}x \right )dx$

Solution:

$\int_{0}^{\frac{\pi }{2}}\left ( a^{2}\cos ^{2}x+b^{2}\sin ^{2}x \right )dx$

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

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(a^{2} \cos ^{2} x+b^{2}-b^{2} \cos ^{2} x\right) d x \\ &=\int_{0}^{\frac{\pi}{3}}\left(a^{2}-b^{2}\right) \cos ^{2} x d x+b^{2} \int d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 2 \cos ^{2} x d x+b^{2} \int_{0}^{\frac{\pi}{2}} 1 d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x) d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned}                                        $\left[\begin{array}{l} \cos ^{2} x+\sin ^{2} x=1 \\ \Rightarrow \sin ^{2} x=1-\cos ^{2} x \end{array}\right]$

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(a^{2} \cos ^{2} x+b^{2}-b^{2} \cos ^{2} x\right) d x \\ &=\int_{0}^{\frac{\pi}{3}}\left(a^{2}-b^{2}\right) \cos ^{2} x d x+b^{2} \int d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 2 \cos ^{2} x d x+b^{2} \int_{0}^{\frac{\pi}{2}} 1 d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x) d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned}                                            $\left [ 2\cos ^{2}\theta =1+\cos 2\theta \right ]$

\begin{aligned} &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 1 d x+\frac{\left(a^{2}-b^{2}\right)^{\frac{\pi}{2}}}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{\left(a^{2}-b^{2}\right)^{\frac{\pi}{2}}}{2} \int_{0}^{\frac{\pi}{3}} \cos 2 x d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned} \quad\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{\left(a^{2}-b^{2}\right)}{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{2}}+\frac{\left(a^{2}-b^{2}\right)}{2}\left[\frac{\sin 2 x}{2}\right]+b^{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{\left(a^{2}-b^{2}\right)}{2}\left[\frac{\pi}{2}-0\right]+\frac{\left(a^{2}-b^{2}\right)}{2 \times 2}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right]+b^{2}\left[\frac{\pi}{2}-0\right] \end{aligned}                $\left [ \sin \pi =\sin 0=0 \right ]$

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