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Question Number 84681 by Power last updated on 15/Mar/20

Commented by mathmax by abdo last updated on 15/Mar/20

$I = \int \frac{15 s i n x + 2 c o s x}{98 s i n x - 7 c o s x} d x \Rightarrow I = \frac{15}{98} \int \frac{s i n x + \frac{2}{15} c o s x}{s i n x - \frac{7}{98} c o s x} d x$ $l e t d e t e r m i n e A = \int \frac{s i n x + a c o s x}{s i n x + b c o s x} d x w e d i t h e c h a n g e m e n t$ $t a n (\frac{x}{2}) = t \Rightarrow A = \int \frac{\frac{2 t}{1 + t^{2}} + a \frac{1 - t^{2}}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} + b \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}}$ $= 2 \int \frac{2 t + a - {a t}^{2}}{(2 t + b - {b t}^{2}) (1 + t^{2})} d t = 2 \int \frac{- {a t}^{2} + 2 t + a}{(- {b t}^{2} + 2 t + b) (t^{2} + 1)} d t$ $= 2 \int \frac{{a t}^{2} - 2 t - a}{(t^{2} + 1) ({b t}^{2} - 2 t - b)} d t l e t d e c o m p o s e$ $F (t) = \frac{{a t}^{2} - 2 t - a}{(t^{2} + 1) ({b t}^{2} - 2 t - b)}$ ${b t}^{2} - 2 t - b = 0 \to Δ^{^{'}} = 1 + b^{2} > 0 \Rightarrow t_{1} = \frac{1 + \sqrt{b^{2} + 1}}{b}$ $t_{2} = \frac{1 - \sqrt{b^{2} + 1}}{b} \Rightarrow F (t) = \frac{{a t}^{2} - 2 t - a}{b (t - t_{1}) (t - t_{2}) (t^{2} + 1)}$ $= \frac{α}{t - t_{1}} + \frac{β}{t - t_{2}} + \frac{λ t + γ}{t^{2} + 1}$ $α = \frac{{a t}_{1}^{2} - 2 t_{1} - a}{b (t_{1} - t_{2}) (t_{1}^{2} + 1)}$ $β = \frac{{a t}_{2}^{2} - 2 t_{2} - a}{b (t_{2} - t_{1}) (t_{2}^{2} + 1)} . . . . b e c o n t i n u e d . . .$
	Answered by TANMAY PANACEA last updated on 15/Mar/20

	$\int \frac{a s i n x + b c o s x}{c s i n x + d c o s x} d x$ $a s i n x + b c o s x = p (c s i n x + d c o s x) + q \times \frac{d}{d x} (c s i n x + d c o s x)$ $s o$ $\int \frac{p (c s i n x + d c o s x) + q \times \frac{d}{d x} (c s i n x + d c o s x)}{c s i n x + d c o s x} d x$ $\int p d x + q \int \frac{d (c s i n x + d c o s x)}{c s i n x + d c o s x}$ $p x + q l n (c s i n x + d c o s x) + C$ $n o w$ $15 s i n x + 2 c o s x = p (98 s i n x - 7 c o s x) + q (98 c o s x + 7 s i n x)$ $98 p + 7 q = 15$ $- 7 p + 98 q = 2 \times 14$ $98 p + 7 q = 15$ $- 98 p + 98 \times 14 q = 28$ $q \times 7 (1 + 196) = 28 q = \frac{28}{7 \times 197} = \frac{4}{197}$ $98 p = 15 - 7 \times \frac{4}{197} = \frac{15 \times 197 - 28}{197}$ $p = (\frac{15 \times 197 - 28}{197 \times 98})$ $p l s c a l c u l a t e p a n d q . . .$
	Commented by Power last updated on 15/Mar/20

	$thanks$
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