Integrate : ∫5sin2x-cosx / 9sin2x+4\(\int\frac{5sin 2x-cos x}{9sin^

1.	Integrate : ∫5sin2x-cosx / 9sin2x+4\(\int\frac{5sin 2x-cos x}{9sin^2x+4}dx\)
Answer» \(\int\frac{5sin 2x-cos x}{9sin^2x+4}dx\) \(=\int\frac{5sin2x}{9sin^2x+4}dx-\int\frac{cos x}{9sin^2x+4}dx\) = \(5\int\frac{dt}{9t+4}-\frac19\int\cfrac{du}{u^2+\frac49}\) (By taking sin²x = t and sin x = u) = \(\frac59\)log\|9t + 4\| - \(\frac16\)tan^-1(3u/2) + c = \(\frac59\) log\|9 sin²x + u\| - \(\frac16\) tan^-1(\(\frac{3sinx}2\)) + c (By putting t = sin²x and u = sin x)

Integrate : ∫5sin2x-cosx / 9sin2x+4\(\int\frac{5sin 2x-cos x}{9sin^2x+4}dx\)

Answer»

\(\int\frac{5sin 2x-cos x}{9sin^2x+4}dx\)

\(=\int\frac{5sin2x}{9sin^2x+4}dx-\int\frac{cos x}{9sin^2x+4}dx\)

= \(5\int\frac{dt}{9t+4}-\frac19\int\cfrac{du}{u^2+\frac49}\) (By taking sin²x = t and sin x = u)

= \(\frac59\)log|9t + 4| - \(\frac16\)tan^-1(3u/2) + c

= \(\frac59\) log|9 sin²x + u| - \(\frac16\) tan^-1(\(\frac{3sinx}2\)) + c

(By putting t = sin²x and u = sin x)