Integral Calculus> please integrate this definite integratio...

please integrate this definite integration at picture bilow

Rajmita sen , 11 Years ago

Grade

1 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

$I = \int_{0}^{\pi /2}\frac{1}{5+4sinx}dx$
$I = \int_{0}^{\pi /2}\frac{1}{5+4sin(\frac{\pi }{2}-x)}dx$
$I = \int_{0}^{\pi /2}\frac{1}{5+4cosx}dx$
$I = \int_{0}^{\pi /2}\frac{1}{5+4(2cos^{2}\frac{x}{2}-1)}dx$
$I = \int_{0}^{\pi /2}\frac{1}{8cos^{2}\frac{x}{2}+1}dx$
$I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{sec^{2}\frac{x}{2}+8}dx$
$I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{tan^{2}\frac{x}{2}+1+8}dx$
$I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{tan^{2}\frac{x}{2}+9}dx$
$t = tan\frac{x}{2}$
$dt = sec^{2}\frac{x}{2}.\frac{1}{2}dx$
$x=0\rightarrow t=0$
$x=\frac{\pi }{2}\rightarrow t=1$
$I = \int_{0}^{1}\frac{2}{t^{2}+9}dt$
$I = \int_{0}^{1}\frac{2}{t^{2}+(3)^{2}}dt$
$I = \frac{2}{3}[tan^{-1}(\frac{t}{3})]_{0}^{1}$
$I = \frac{2}{3}tan^{-1}(\frac{1}{3})$

Last Activity: 11 Years ago