First slide

Methods of integration

Question

If the integral $\int \frac{5 \tan x}{\tan x - 2} d x = x +$ $a \log | \sin x - 2 \cos x | + ∣ K$ then a is equal to

Moderate

A

$- 2$
B

1
C

2
D

-1

Solution

$I = \int \frac{5 \tan x}{\tan x - 2} d x = \int \frac{5 \sin x}{\sin - 2 \cos x} d x$
Let $5 \sin x = A (\sin x - 2 \cos x) + B (\cos x + 2 \sin x)$
$Equating coefficients of \sin x and \cos x, we get$
$5 = A + 2 B$ and $0 = - 2 A + B$
$\begin{array}{l} \Rightarrow A = 1, B = 2 \\ ∴ I = \int \frac{(\sin x - 2 \cos x) + 2 (\cos x + 2 \sin x)}{\sin x - 2 \cos x} d x \\ = x + 2 \log | \sin x - 2 \cos x | + K \\ ∴ a = 2 \end{array}$

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