The value of the integral
`int_(0)^(pi//2)(3sqrt(costheta))/((sqrt(costheta)+sqrt(sintheta))^5)d theta` equals ............

Answer

`0.5`

Explanation

Ket Idea : Use property `int _(0)^(a)f(x)dx=int _(0)^(a)f(a-x)dx`
The given intergral
`I= int_(0)^(pi//2)(3sqrt(cos theta))/((sqrt(cos theta) +sqrt(sin theta ))^5)d theta` ......(i)
`I= int_(0)^(pi//2)(3sqrt(sin theta))/((sqrt(sin theta) +sqrt(cos theta ))^5)d theta` ......(ii)
[Using the property `int _(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`]
Now, on adding integrals (1) and (ii), we get
`2I= int_(0)^(pi//2)(3)/((sqrt(sintheta) +sqrt(costheta ))^4)d theta`
`=int _(0)^(pi//2)(3sec^2 theta)/(1+sqrt(tan theta ))^4d theta`
Now, let `tan theta =t^2rArr sec^2 theta = 2t dt`
and at `theta=(pi)/(2),t to oo`
and at `theta= 0,t to 0`
So, `2I=int _(0)^(oo) (6t dt)/(1+t)^4=6int_(0)^(oo)(t+I-1)/(t+1)^4=dt `
`rArr I=3[int_(0)^(oo)(dt)/(t-1)^3-int_(0)^(oo)(dt)/(t+I)^4]=3[-(1)/(2(t+1)^2)+(1)/(3(t+1)^3)]_(0)^(oo)`
`rArr I=3 [(1)/(2)-(1)/(3)]=3((1)/(6))=(1)/(2)rArr I=0.5`.