CUET Mathematics 2024 - _0^/2 1- xcosec x+ x d x= | PYQs + Solutions

CUET Mathematics 2024

Calculus

Integrals

Medium

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

0

\frac{\pi}{4}

\infty

\frac{\pi}{12}

✅ Correct Option: 1

\int_{0}^{\frac{\pi}{2}} \dfrac{1-\cot x}{\operatorname{cosec} x+\cos x} dx

Let's simplify the integrand by converting everything in terms of

\sin x

and

\cos x

Remember that

\cot x = \dfrac{\cos x}{\sin x}

and

\operatorname{cosec} x = \dfrac{1}{\sin x}

Substituting these into our integrand:

\dfrac{1-\cot x}{\operatorname{cosec} x+\cos x} = \dfrac{1-\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x}+\cos x}

Simplifying the numerator:

1-\dfrac{\cos x}{\sin x} = \dfrac{\sin x - \cos x}{\sin x}

Now our integrand becomes:

\dfrac{\dfrac{\sin x - \cos x}{\sin x}}{\dfrac{1}{\sin x}+\cos x} = \dfrac{\sin x - \cos x}{\sin x} \cdot \dfrac{\sin x}{1+\sin x \cos x} = \dfrac{\sin x - \cos x}{1+\sin x \cos x}

Let's use the substitution

u = \dfrac{\pi}{2} - x

, then:

\newline

du = -dx

\newline

- When

x = 0

u = \dfrac{\pi}{2}

\newline

- When

x = \dfrac{\pi}{2}

u = 0

Under this substitution:

\newline

\sin u = \cos x

\newline

\cos u = \sin x

Applying the substitution to the integral:

\int_{0}^{\frac{\pi}{2}} \dfrac{\sin x - \cos x}{1+\sin x \cos x} dx = \int_{\frac{\pi}{2}}^{0} \dfrac{\cos u - \sin u}{1+\cos u \sin u} (-du) = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin u - \cos u}{1+\sin u \cos u} du

Let's call our original integral

I

I = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x - \cos x}{1+\sin x \cos x} dx

From our substitution, we also know:

I = \int_{0}^{\frac{\pi}{2}} \dfrac{\cos u - \sin u}{1+\sin u \cos u} du

Renaming

u

back to

x

in the second integral:

I = \int_{0}^{\frac{\pi}{2}} \dfrac{\cos x - \sin x}{1+\sin x \cos x} dx

Adding these equal values:

2I = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x - \cos x}{1+\sin x \cos x} dx + \int_{0}^{\frac{\pi}{2}} \dfrac{\cos x - \sin x}{1+\sin x \cos x} dx

2I = \int_{0}^{\frac{\pi}{2}} \dfrac{(\sin x - \cos x) + (\cos x - \sin x)}{1+\sin x \cos x} dx = \int_{0}^{\frac{\pi}{2}} \dfrac{0}{1+\sin x \cos x} dx = 0

Therefore,

I = 0

\int_{0}^{\frac{\pi}{2}} \dfrac{1-\cot x}{\operatorname{cosec} x+\cos x} dx = 0