Explain Solution R .D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 7 Maths Textbook Solution.

\begin{aligned} &\text { Answer }:-\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+C\\ &\text { Hint :- Use substitution method to solve this integral. }\\ &\text { Given :- } \int \sec ^{4} 2 x d x\\ &\text { Sol: - Let } I=\int \sec ^{4} 2 x d x\\ &\text { Re-writing, } I=\int \sec ^{2} 2 x \cdot \sec ^{2} 2 x d x \end{aligned}

\begin{aligned} &\left.I=\int\left(1+\tan ^{2} 2 x\right) \sec ^{2} 2 x d x\: \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right) \\ &I=\int\left(\sec ^{2} 2 x+\tan ^{2} 2 x \cdot \sec ^{2} 2 x\right) d x \end{aligned}

\begin{aligned} &\text { Substitute } \tan 2 \mathrm{x}=\mathrm{t}\\ &\sec ^{2} 2 x .2 d x=d t, \text { then } \end{aligned}

\begin{aligned} &\mathrm{I}=\int \sec ^{2} 2 x \mathrm{~d} x+\int \tan ^{2} 2 x \cdot \sec ^{2} 2 x \mathrm{dx} \\ &=\int \sec ^{2} 2 x \mathrm{~d} x+\int t^{2} \cdot \frac{d t}{2} \: \: \: \: \: \: \: \: \: \: \quad(\text { since }, \tan 2 x=t) \end{aligned}

$\inline =\frac{\tan 2 x}{2}+\frac{1}{2} \frac{t^{2+1}}{2+1}+C$                                                            $\inline \text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \sec ^{2} d x=\tan x+\mathrm{c}$

\inline \begin{aligned} &=\frac{\tan 2 x}{2}+\frac{1}{2} \cdot \frac{t^{3}}{3}+C \\ &\left.=\frac{\tan 2 x}{2}+\frac{1}{6} \cdot \tan ^{3} 2 x+C \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } t=\tan 2 x\right) \end{aligned}