Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 6 Maths Textbook Solution.

Answer: $\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+C$

Hint: Use substitution method to solve this integral.

Given:$\int \sqrt{\tan x} \cdot \sec ^{4} x d x$

Solution:Let, $I=\int \sqrt{\tan x} \cdot \sec ^{4} x d x$

Re-Write $\mathrm{I}=\int \sqrt{\tan x} \cdot \sec ^{2} x \cdot \sec ^{2} x d x$

$I=\int \sqrt{\tan x}\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad\left(\because \sec ^{2} x-\tan ^{2}=1\right)$

$I=\int\left(\sqrt{\tan x}+\sqrt{\tan x} \cdot \tan ^{2} x\right) \sec ^{2} x d x$

$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{1}{2}+2} x\right) \cdot \sec ^{2} x d x$

$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \cdot \sec ^{2} x d x$

$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{t} \text { , then }$

$\begin{array}{ll} \mathrm{I}=\int\left(t^{\frac{1}{2}}+t^{5 / 2}\right) d t \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & (\because \tan \mathrm{x}=\mathrm{t}) \end{array}$

$=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{5}{2}}+1}{\frac{5}{2}+1}+C \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$

$=\frac{\tan ^{3 / 2} x}{3 / 2}+\frac{\tan ^{7 / 2} x}{7 / 2}+\mathrm{C} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because \mathrm{t}=\tan \mathrm{x})$

$=\frac{2}{3} \tan ^{3 / 2} x+\frac{2}{7} \tan ^{7 / 2} x+C$