First slide

Introduction to integration

Question

If $\int \frac{\cos^{4} x}{\sin^{2} x} dx = Acot x + Bsin 2 x + \frac{C}{2} x + D$ , then

Easy

A

$A = - 2, B = \frac{1}{4}$
B

$B = - \frac{1}{4}, C = - 3$
C

$B = \frac{1}{4}, C = - 3$
D

None of these

Solution

We have, 1
$\begin{array}{l} = \int \frac{\cos^{4} x}{\sin^{2} x} dx \\ = \int \frac{{(1 - \sin^{2} x)}^{2}}{\sin^{2} x} dx \\ = \int ({cosec}^{2} x + \sin^{2} x - 2) dx \\ = - \cot x + \frac{1}{2} (x - \frac{\sin 2 x}{2}) - 2 x + D \\ = - \cot x - \frac{1}{4} \sin 2 x - \frac{3}{2} x + D \end{array}$
$∴ A = - 1, B = - \frac{1}{4}, C = - 3$

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