Question

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 \(a^{10}+b^{5}=\cdots\)

• A. \(16\)
• B. \(64\)
• C. \(36\)
• D. \(100\)

Hint:

Substitute u equal to the square root of tan x and reduce to a standard form using u minus 1 over u.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Write \(\sqrt{\tan x}+\sqrt{\cot x}=\dfrac{\tan x+1}{\sqrt{\tan x}}\).

Step 2: Substitute
Let \(u=\sqrt{\tan x}\). Then \(dx=\dfrac{2u\,du}{1+u^{4}}\).
The integral becomes \(2\int\dfrac{u^{2}+1}{u^{4}+1}du=2\int\dfrac{1+1/u^{2}}{u^{2}+1/u^{2}}du\).

Step 3: Use w
Let \(w=u-\dfrac1u\). Then \(dw=\left(1+\dfrac1{u^{2}}\right)du\) and \(u^{2}+\dfrac1{u^{2}}=w^{2}+2\).
Integral \(=2\int\dfrac{dw}{w^{2}+2}=\sqrt2\tan^{-1}\dfrac{w}{\sqrt2}\).

Step 4: Compare
\(\dfrac{w}{\sqrt2}=\dfrac{\tan x-1}{\sqrt{2\tan x}}\).
So \(a=\sqrt2\) and \(b=2\).

Step 5: Calculation
\(a^{10}+b^{5}=2^{5}+2^{5}=64\).

Final Answer:
The value of a to the tenth plus b to the fifth is 64. \[ \boxed{\text{(B) }64} \]