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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.

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

Answers (1)

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)

 

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