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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short answers Question 61

Answers (1)

Answer: \tan x-\cot x+c

Hints: You must know about the rules of integration of trigonometric and inverse trigonometric functions.

Given: \int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x} d x \\

Solution:

\begin{aligned} &\int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ &\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x \end{aligned}

\begin{aligned} &\int \frac{\sin ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ &=\int \frac{1}{\cos ^{2} x} d x+\int \frac{1}{\sin ^{2} x} d x \\ &=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x \\ &=\tan x-\cot x+c \end{aligned}

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