Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 81 Maths Textbook Solution.

$\frac{\sec ^{7} x}{7}-\frac{2 \sec ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$

Hint:

You must know about integration of tan x & sec x.

Given:

$\int \tan ^{5} x \sec ^{3} x d x$

Solution:

$\text { let } \sec x=t$

$\sec x \tan x d x=d t$

$\text { Now, } \int \tan ^{4} x \sec ^{2} x(\sec x \tan x d x) \ldots \ldots \ldots . \text { (1) }$

$\int \tan ^{4} x \sec ^{2} x d t$

$\text { put the value of } \sec x=\operatorname{tin}(1)$

$\int \tan ^{2} x \cdot \tan ^{2} x \cdot \sec ^{2} x d t$

$\int\left(\left(\sec ^{2} x-1\right)\left(\sec ^{2} x-1\right) \cdot \sec ^{2} x\right) d t$

$\int\left(t^{2}-1\right)\left(t^{2}-1\right) \cdot t^{2} d t$

$\int\left[t^{4}+(1)^{2}-2\left(t^{2}\right)(1) .\right] t^{2} d t$

$\int\left(t^{4}+1-2 t^{2}\right) t^{2} d t$

$\int t^{6} d t-\int 2 t^{4} d t+\int t^{2} d t$

$=\frac{t^{7}}{7}-\frac{2 t^{5}}{5}+\frac{t^{3}}{3}+c$

$=\frac{\sec ^{7} x}{7}-\frac{2 \sec ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$