If the integral \int \frac{5 tanx}{tanx - 2} \textrm{ } dx = x + a ln \left|\right. sinx - 2 cosx \left|\right. + k then a is equal to :

Engineering

Mathematics

Integration of Trigonometric Functions

Question

If the integral $\int \frac{5 tanx}{tanx - 2} dx = x + a \ln | sinx - 2 cosx | + k$ then a is equal to :

–1

–2

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Differentiating both sides,

$\frac{5 tanx}{tanx - 2} = 1 + \frac{a}{sinx - 2 cosx} (cosx + 2 sinx)$

$\frac{5 sinx}{sinx - 2 cosx} - \frac{a (cosx + 2 sinx)}{sinx - 2 cosx} = 1$

5sin x – a (cos x + 2sin x) = sin x – 2cos x

4sin x + 2cos x = a (cos x + 2sin x)

$a = \frac{2 (2 sinx + cosx)}{cosx + 2 sinx} \Rightarrow a = 2$

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