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GUJCET 2026 Math Paper

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Question : 3 of 40

Marks: +1, -0

\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}} =

. . . . . . .

$\frac{\pi}{12}$
$\frac{\pi}{6}$
$0$
$1$

Solution:

Concept:

Use the definite integral property

\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx

to simplify symmetric limits.

Explanation:

Let

I=\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\cot x}}

.

Since

\sqrt{\cot x}=\frac{\sqrt{\cos x}}{\sqrt{\sin x}}

, we get

I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx

.

Here

a=\frac{\pi}{6}

and

b=\frac{\pi}{3}

, so

a+b=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}

.

Apply the property with

x\to \frac{\pi}{2}-x

:

I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}}\,dx

.

Using

\sin(\frac{\pi}{2}-x)=\cos x

and

\cos(\frac{\pi}{2}-x)=\sin x

:

I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx

.

Add the two forms of

I

:

2I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx

.

2I=\int_{\pi/6}^{\pi/3}1\,dx=[x]_{\pi/6}^{\pi/3}

.

2I=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}

.

Therefore,

I=\frac{\pi}{12}

.

Answer:

\frac{\pi}{12}

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