If \(\int (\sqrt{\tan x}+\sqrt{\cot x})dx=a\tan ^{-1}\left(\frac{\tan x-1}{\sqrt{b\tan x}}\right)+c\), for \(x\in \left(0,\frac{\pi }{2}\right)\), then \(a^{10}+b^{5}=\cdots\)
Show Hint
Substitute u equal to the square root of tan x and reduce to a standard form using u minus 1 over u.
Step 1: Plan
Differentiate the given form with $t=\tan x$ and match it with the integrand $\dfrac{t+1}{\sqrt t}$.
Step 2: Differentiate
Let $w=\dfrac{t-1}{\sqrt{bt}}$. Then $\dfrac{dw}{dx}=\dfrac{1}{2\sqrt b}\cdot\dfrac{(t+1)(1+t^{2})}{t^{3/2}}$ and $1+w^{2}=\dfrac{bt+(t-1)^{2}}{bt}$.\nFor $b=2$, $1+w^{2}=\dfrac{t^{2}+1}{2t}$.
Step 3: Match
Then $\dfrac{d}{dx}\left[a\tan^{-1}w\right]=\dfrac{a}{\sqrt2}\cdot\dfrac{t+1}{\sqrt t}$.\nThis equals $\dfrac{t+1}{\sqrt t}$ only when $a=\sqrt2$ with $b=2$.
Step 4: Compute
$a^{10}=(\sqrt2)^{10}=32$ and $b^{5}=2^{5}=32$, so the sum is $64$.
Final Answer:
With a equal to root 2 and b equal to 2 the sum is 64.
\[ \boxed{\text{(B) }64} \]