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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 10 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 10 Maths Textbook Solution.

Answers (1)

Answer: -\frac{1}{6} \cot ^{6} x-\frac{1}{8} \cot ^{8} x+C

Hint :-  Use substitution method to solve this integral.

Given:\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x

Solution:Let, \mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x

Re-Write,\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x

\mathrm{I}=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x            \text { (if, } \left.\operatorname{cosec}^{2} x-\cot ^{2} x=1\right)

\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{5} x \cdot \cot ^{2}\right) \operatorname{cosec}^{2} x d x

\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x

Substitute,  cotx = t

-\operatorname{cosec}^{2} x d x=d t, \text { then }

I=\int\left(t^{5}+t^{7}\right) \operatorname{cosec}^{2} x \cdot \frac{d t}{-\operatorname{cosec}^{2} x} \mathrm{dx} \: \: \: \: \: \: \: \: \quad(\text { as } \cot x=t)

\begin{aligned} &=-\int\left(t^{5}+t^{7}\right) \mathrm{dt} \\ &=-\int t^{5} d t-\int t^{7} \mathrm{dt} \end{aligned}

=-\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}+\mathrm{c}\right)

=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+C

=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+\mathrm{C}                    (\text { if, } t=\cot x)

 

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