Consider sin^3xsin(x+α) =sin^3x[sinxcosα+cosxsinα] #160; #160; #160; #160; #160; =sin^4xcosα+sin^3xcosxsinα =sin^4xcosα+sin^3xc ...

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Integration by Partial Fractions

∫1√sin3xsin x...

Question

$\int \frac{1}{\sqrt{{sin}^{3} x sin (x + α)}} d x$ .

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Solution

Consider ${sin}^{3} x sin (x + α)$
$= {sin}^{3} x [sin x cos α + cos x sin α]$
$= {sin}^{4} x cos α + {sin}^{3} x cos x sin α$
$= {sin}^{4} x cos α + {sin}^{3} x cos x sin α \times \frac{sin x}{sin x}$
$= {sin}^{4} x [cos α + \frac{cos x}{sin x} sin α]$
$∴ {sin}^{3} x sin (x + α) = {sin}^{4} x [cos α + cot x sin α]$ ...... $(1)$
$∴ {sin}^{3} x sin (x + α) = {sin}^{4} x (cos α + cot x sin α)$
Now, $\int \frac{d x}{\sqrt{{sin}^{3} x sin (x + α)}}$
$= \int \frac{d x}{\sqrt{{sin}^{4} x (cos α + cot x . sin α)}}$
$= \int \frac{d x}{{sin}^{2} x \sqrt{(cos α + cot x . sin α)}}$
Let $cos α + cot x sin α = t$
Differentiating w.r.t $x$ we get
$0 - sin α {csc}^{2} x d x = d t$
$\Rightarrow d x = \frac{- d t}{sin α {csc}^{2} x}$
Now our equation becomes $= \int \frac{d x}{{sin}^{2} x \sqrt{(cos α + cot x . sin α)}}$
$= \int \frac{1}{{sin}^{2} x \sqrt{t}} \times \frac{1}{- sin α} \times {sin}^{2} x d t$
$= \int \frac{1}{{sin}^{2} x \sqrt{t}} \times \frac{1}{- sin α} \times {sin}^{2} x d t$
$= \frac{- 1}{sin α} \int \frac{d t}{\sqrt{t}}$
$= \frac{- 1}{sin α} \int t^{\frac{- 1}{2}} d t$
$= \frac{- 1}{sin α} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{t^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + c ⎤ ⎥ ⎥ ⎥ ⎦$
$= \frac{- 1}{sin α} [2 \sqrt{t} + c]$
Putting the value of $t = cos α + cot x sin α$
$= \frac{- 1}{sin α} [2 \sqrt{cos α + cot x sin α} + c]$
$= \frac{- 2}{sin α} [\sqrt{cos α + cot x sin α} + c]$
From $(1)$
$\frac{{sin}^{3} x sin (x + α)}{{sin}^{4} x} = cos α + cot x sin α$
$\frac{sin (x + α)}{sin x} = cos α + cot x sin α$
$= \frac{- 2}{sin α} [\sqrt{cos α + cot x sin α} + c]$
$= \frac{- 2}{sin α} \sqrt{\frac{sin (x + α)}{sin x}} + c$

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