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

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

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:

• $du = -dx$
  • When $x = 0$, $u = \dfrac{\pi}{2}$
  • When $x = \dfrac{\pi}{2}$, $u = 0$

Under this substitution:

• $\sin u = \cos x$
  • $\cos u = \sin x$

---

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, $I = 0$

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