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#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.11 Question 3 Maths Textbook Solution.

Answer: $\frac{1}{6} \tan ^{6} x+\frac{1}{8} \tan ^{8} x+C$

Hint: Use substitution method to solve this integral.

Given: $\int \tan ^{5} x \cdot \sec ^{4} x d x$

Solution: Let,$\mathrm{I}=\int \tan ^{5} x \sec ^{4} x d x$

Re-Write $I=\int \tan ^{5} x \sec ^{2} \cdot \sec ^{2} x d x$

$\left.I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right)$

$\mathrm{I}=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$

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

$\left.I=\int\left(\mathrm{t}^{5}+t^{7}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t}\right)$

$=\int\left(\mathrm{t}^{5}+t^{7}\right) d t=\int \mathrm{t}^{5} d t+\int t^{7} d t$

$=\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C$                            $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right)$......(i)

$\mathrm{I}=\frac{t^{6}}{6}+\frac{t^{8}}{8}+\mathrm{C}$

$\left.\mathrm{I}=\frac{\tan ^{6} x}{6}+\frac{\tan ^{8} x}{8}+\mathrm{C}\: \: \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\tan \mathrm{x}\right)$