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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 32 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 32 Maths Textbbok Solution.

Answers (1)

Answer:

-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+c

Given:

\int \cot ^{5} x d x

Hint:

To solve the given statement we will split into \cot ^{3} x \cot ^{2} x

Solution: 

   \int \cot ^{3} x \cot ^{2} x d x

I=\int \cot ^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x

    =\int \cot ^{3} x \operatorname{cosec}^{2} x d x-\int \cot x\left(\operatorname{cosec}^{2} x-1\right) d x

    =\int \cot ^{3} x \operatorname{cosec}^{2} x d x-\int \cot x \operatorname{cosec}^{2} x d x+\int \cot x d x

    =-\int p^{3} d p+\int p d p+\int \cot x d x \quad[\because \cot x=p, \operatorname{cosec} x d x=d p]

    =-\frac{p^{4}}{4}+\frac{p^{2}}{2}+\log \mid(\sin x)+c

     =-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+c

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