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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 71 Maths Textbbok Solution.

Answers (1)

Answer:

-\frac{\cot 2 x}{2}-\frac{\cot ^{8} 2 x}{6}+c

Hint:

To solve the given question we will split the cosec? 2x into cosec² 2x cosec² 2x.

Given:

\int \operatorname{cosec}^{4} 2 x d x

Solution:

=\int \operatorname{cosec}^{4} 2 x d x

=\int \operatorname{cosec}^{2} 2 x \operatorname{cosec}^{2} 2 x d x

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

p u t \cot 2 x=t=>-\operatorname{cosec}^{2} 2 x \cdot 2 d x=d t

=\frac{-1}{2} \int\left(1+t^{2}\right) \cdot d t

=-\frac{1}{2} t-\frac{t^{3}}{6}+c

=-\frac{\cot 2 x}{2}-\frac{\cot ^{3} 2 x}{6}+c

 

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