#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 67 Maths Textbook Solution.

Answer: $\frac{\left ( a^{2}+b^{2} \right )\pi }{8}+\frac{\left ( a^{2}-b^{2} \right )}{4}$

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

Given: $\int_{0}^{\frac{\pi}{4}}\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right) d x$

Solution:

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

$=a^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} 1 d x+\left(b^{2}-a^{2}\right) \int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\cos 2 \theta=2 \sin ^{2} \theta\right] \$

$=a^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} 1 d x+\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\$

$=\left(a^{2}+\frac{\left(b^{2}-a^{2}\right)}{2}\right)_{0}^{\frac{\pi}{4}} x d x-\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\$

$=\left(\frac{2 a^{2}+\left(b^{2}-a^{2}\right)}{2}\right)\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{4}}-\frac{\left(b^{2}-a^{2}\right)}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}} \\$

$=\left(\frac{a^{2}+b^{2}}{2}\right)[x]_{0}^{\frac{\pi}{4}}-\frac{\left(b^{2}-a^{2}\right)}{2 \times 2}\left[\sin 2 \times \frac{\pi}{4}-\sin 2 \times 0\right] \\$

$=\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{\pi}{4}-0\right)-\frac{\left(b^{2}-a^{2}\right)}{4}\left[\sin \left(\frac{\pi}{2}\right)-\sin 0\right]$

$\left [ \sin \frac{\pi }{2}=1,\sin 0=0 \right ]$

\begin{aligned} &=\left(\frac{a^{2}+b^{2}}{2}\right) \frac{\pi}{4}-\frac{\left(b^{2}-a^{2}\right)}{4} \sin (1-0) \\ &=\left(\frac{a^{2}+b^{2}}{2}\right) \frac{\pi}{4}+\left(\frac{a^{2}-b^{2}}{4}\right) \end{aligned}