Evaluate :
`(i) int_(0)^(2pi) {sin(sinx)+sin(cosx)}dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{2\pi} \left( \sin(\sin x) + \sin(\cos x) \right) dx, \]

we can utilize the property of definite integrals. This property states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = 2\pi \), so we can rewrite the integral as:

\[ I = \int_{0}^{2\pi} \left( \sin(\sin(2\pi - x)) + \sin(\cos(2\pi - x)) \right) dx. \]

Now, we simplify the terms inside the integral:

1. **Simplifying \( \sin(\sin(2\pi - x)) \)**:
\[ \sin(2\pi - x) = -\sin(x). \]
Therefore,
\[ \sin(\sin(2\pi - x)) = \sin(-\sin(x)) = -\sin(\sin(x)). \]

2. **Simplifying \( \sin(\cos(2\pi - x)) \)**:
\[ \cos(2\pi - x) = \cos(x). \]
Thus,
\[ \sin(\cos(2\pi - x)) = \sin(\cos(x)). \]

Putting these together, we have:

\[ I = \int_{0}^{2\pi} \left( -\sin(\sin x) + \sin(\cos x) \right) dx. \]

Now, we denote this new integral as \( J \):

\[ J = \int_{0}^{2\pi} \left( -\sin(\sin x) + \sin(\cos x) \right) dx. \]

Now, we can add the two integrals \( I \) and \( J \):

\[ I + J = \int_{0}^{2\pi} \left( \sin(\sin x) + \sin(\cos x) - \sin(\sin x) + \sin(\cos x) \right) dx. \]

This simplifies to:

\[ I + J = \int_{0}^{2\pi} 2\sin(\cos x) \, dx. \]

Since \( J = -I \), we have:

\[ I - I = \int_{0}^{2\pi} 2\sin(\cos x) \, dx. \]

Thus,

\[ 2I = \int_{0}^{2\pi} 2\sin(\cos x) \, dx. \]

Now, we can simplify this further:

\[ I = \int_{0}^{2\pi} \sin(\cos x) \, dx. \]

Next, we can evaluate \( \int_{0}^{2\pi} \sin(\cos x) \, dx \) using the property of symmetry. The function \( \sin(\cos x) \) is periodic and symmetric about \( \pi \).

Thus, we can conclude that:

\[ \int_{0}^{2\pi} \sin(\cos x) \, dx = 0. \]

Finally, we have:

\[ I = 0. \]

**Final Answer:**

\[ \int_{0}^{2\pi} \left( \sin(\sin x) + \sin(\cos x) \right) dx = 0. \]

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