Question

If $I=\int \frac{\sin x{\left(\cos x\right)}^{-5/2}dx}{\sqrt{\sin x+3\cos x}+\sqrt{\sin x+4\cos x}}$$=\frac{A}{1550}({(\tan x+4)}^{5/2}-{(\tan x+3)}^{3/2})-\frac{2}{3}[4{(\tan x+4)}^{3/2}-3{(\tan x+3)}^{3/2}]+C$,

then the value of A is equal to

Answer 1

$I=\int \frac{\sin x{\left(\cos x\right)}^{-5/2}dx}{\sqrt{\sin x+3\cos x}+\sqrt{\sin x+4\cos x}}=\int \frac{\sin x{\left(\cos x\right)}^{-3}}{\sqrt{\tan x+3}+\sqrt{\tan x+4}}$
Let $\tan x=t$
$I=\int \frac{t}{\sqrt{t+3}+\sqrt{t+4}}dt=I_4-I_3$
Where $I_a=\int t\sqrt{t+a}dt$
Susbtituting $t+a=u^2$
$I_a=\int (u^2-a)u\left(2u\right)du=\frac{2u^5}{5}-\frac{2a{u}^3}{3}=\frac{2{(t+a)}^{5/2}}{5}-\frac{2a{(t+a)}^{3/2}}{3}$
Therefore
$I=\frac{2}{5}({(\tan x+4)}^{5/2}-{(\tan x+3)}^{5/2})-\frac{2}{3}(4{(\tan x+4)}^{3/2}-3{(\tan x+3)}^{3/2})+c$
Therefore, $A=620$