A:∫14+5sinx dx=13log2 tan(x/2)+12 tan(x/2)+4+cR:If 0<a<b, then ∫dxa+bsinx=1b2−a2log a tan x/2+b-b2-a2a tan x/2+b+b2-a2+c

I=∫dx4+5sin⁡x tanx2=t⇒dx=2dt1+t2 sin⁡x=2t1+t2I=∫2dt4t2+10+4=12∫dtt2+5t2+1I=12∫dtt+542−916=12∫dtt+542−342=12×1234log⁡t+12t+2=13log⁡2tanx2+12tanx2+2+C.

A:∫14+5sinx dx=13log2 tan(x/2)+12 tan(x/2)+4+cR:If 0<a<b, then ∫dxa+bsinx=1b2−a2log a tan x/2+b-b2-a2a tan x/2+b+b2-a2+c

I=∫dx4+5sin⁡x tanx2=t⇒dx=2dt1+t2 sin⁡x=2t1+t2I=∫2dt4t2+10+4=12∫dtt2+5t2+1I=12∫dtt+542−916=12∫dtt+542−342=12×1234log⁡t+12t+2=13log⁡2tanx2+12tanx2+2+C.

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$A : \int \frac{1}{4 + 5 \sin x} d x = \frac{1}{3} \log |\frac{2 \tan (x / 2) + 1}{2 \tan (x / 2) + 4}| + c$
$R : I f 0 < a < b, t h e n$ $\int \frac{d x}{a + b \sin x} = \frac{1}{\sqrt{b^{2} - a^{2}}} \log$ $|\frac{a \tan (x / 2) + b - \sqrt{b^{2} - a^{2}}}{a \tan (x / 2) + b + \sqrt{b^{2} - a^{2}}}| + c$

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Detailed Solution

$\begin{array}{l} I = \int \frac{d x}{4 + 5 \sin x} \tan \frac{x}{2} = t \Rightarrow d x = \frac{2 d t}{1 + t^{2}} \sin x = \frac{2 t}{1 + t^{2}} \end{array}$
$\begin{array}{l} I = \int \frac{2 d t}{4 t^{2} + 10 + 4} = \frac{1}{2} \int \frac{d t}{t^{2} + \frac{5 t}{2} + 1} \\ I = \frac{1}{2} \int \frac{d t}{{(t + \frac{5}{4})}^{2} - \frac{9}{16}} \\ = \frac{1}{2} \int \frac{d t}{{(t + \frac{5}{4})}^{2} - {(\frac{3}{4})}^{2}} = \frac{1}{2} \times [\frac{1}{2 (\frac{3}{4})} \log (\frac{t + \frac{1}{2}}{t + 2})] \\ = \frac{1}{3} \log (\frac{2 \tan \frac{x}{2} + 1}{2 (\tan \frac{x}{2} + 2)}) + C . \end{array}$