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Question Number 156761 by gsk2684 last updated on 15/Oct/21

p r o v e t h a t

\int \frac{a + b \sin x}{{(b + a \sin x)}^{2}} d x = \frac{- \cos x}{b + a \sin x}

Answered by cortano last updated on 15/Oct/21

\frac{d}{dx} \int \frac{a + bsin x}{{(b + asin x)}^{2}} dx = \frac{d}{dx} (\frac{- \cos x}{b + asin x})

\frac{a + bsin x}{{(b + asin x)}^{2}} = \frac{\sin x (b + asin x) + \cos x (acos x)}{{(b + asin x)}^{2}}

= \frac{bsin x + asin^{2} x + acos^{2} x}{{(b + asin x)}^{2}}

= \frac{a + bsin x}{{(b + asin x)}^{2}} .

Commented by gsk2684 last updated on 15/Oct/21

r e s u l t i s v e r i f i e d . g o o d

Answered by puissant last updated on 15/Oct/21

D = \int \frac{a + b s i n x}{{(b + a s i n x)}^{2}} d x = \int \frac{(a + b s i n x)}{{(b + a s i n x)}^{2}} ∙ \frac{{s e c}^{2} x}{{s e c}^{2} x} d x

= \int \frac{a {s e c}^{2} x + b t a n x s e c x}{{(b s e c x + a t a n x)}^{2}} d x

u = b s e c x + a t a n x \to d u = (b s e c x t a n x + a {s e c}^{2} x) d x

\Rightarrow D = \int \frac{d t}{t^{2}} = \frac{- 1}{t} + C . .

\Rightarrow D = \frac{- 1}{b s e c x + a t a n x} + C

\Rightarrow D = \frac{- 1}{\frac{b}{c o s x} + \frac{a s i n x}{c o s x}} + C = \frac{- c o s x}{b + a s i n x} + C

∴∵ D = \frac{- c o s x}{b + a s i n x} + C . .

Commented by gsk2684 last updated on 15/Oct/21

t h a n k y o u

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