Question

`int_0^(pi/2) dx/(5+4cosx)`?

Answer 1

`2/3arc tan(1/3).`

Explanation:

Let, `I=int_0^(pi/2) dx/(5+4cosx).`

The Proper Substn. for this type of Integrals is, `tan(x/2)=t,` so

that,

`1/2sec^2(x/2)dx=dt, or, dx=(2dt)/sec^2(x/2)=(2dt)/(1+tan^2(x/2)),`

i.e., `dx=(2dt)/(1+t^2).`

When, `x=0, t=tan0=0, &, x=pi/2, t=tan(pi/4)=1.`

Further, `5+4cosx=5+4{(1-tan^2(x/2))/(1+tan^2(x/2))}=5+4((1-t^2)/(1+t^2))`

`=(9+t^2)/(1+t^2).`

`:. I=int_0^1 {(2dt)/(1+t^2)}/{(9+t^2)/(1+t^2)}=2int_0^1dt/(t^2+3^2),`

`=2*1/3[arc tan (t/3)]_0^1,`

`=2/3[arc tan(1/3)-0],`

`:. I=2/3arc tan(1/3).`

Enjoy Maths.!