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Sin(A+B)Sin(A-B)

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Question

Evaluate: $\int \frac{sin x + cos x}{9 + 16 sin 2 x} d x$

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Solution

$\int \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x$

$s i n 2 x = 1 - {(s i n x - c o s x)}^{2}$

$= \int \frac{s i n x c o s x d x}{9 + 16 [1 - {(s i n x - c o s x)}^{2}]}$

$\Rightarrow \int \frac{s i n x + c o s x}{25 - 16 {(s i n x - c o s x)}^{2}}$
Putting $s i n x - c o s x = m$
$(s i n x + c o s x) d x = d m$

$= \int \frac{d m}{25 - 16 m^{2}}$

$= \int \frac{d m}{(5 + 4 m) (5 - 4 m)} = - \int \frac{d m}{(5 + 4 m) (4 m - 5)}$

$= \frac{- 1}{16} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \int \frac{d m}{(m - \frac{5}{4})} - \int \frac{d m}{(m + \frac{5}{4})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ \times \frac{2}{5}$

$= \frac{- 1}{40} [l n | m - 5 / 4 | - l n ∣ ∣ ∣ m + \frac{5}{4} ∣ ∣ ∣]$

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Sin(A+B)Sin(A-B)

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