Question

\[\int\limits_a^b \cos\ x\ dx\]

Answer 1

\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]

\[\text{Here }a = a, b = b, f\left( x \right) = \cos x, h = \frac{b - a}{n}\]
Therefore,
\[I = \int_a^b \cos x d x\]
\[ = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + . . . + f\left\{ a + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \cos\left( a \right) + \cos\left( a + h \right) + . . . + \cos\left\{ a + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \frac{\cos\left\{ a + \left( n - 1 \right)\frac{h}{2} \right\}\sin\frac{nh}{2}}{\sin\frac{h}{2}} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\frac{h}{2}}{\sin\frac{h}{2}}2\cos\left( a + \frac{b - a}{2} - \frac{h}{2} \right) \sin\left( \frac{b - a}{2} \right) \right] ..............\left(\text{Using }nh = b - a \right)\]
\[ = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin\frac{h}{2}} \times \lim_{h \to 0} 2\cos\left( \frac{a + b}{2} - \frac{h}{2} \right)\sin\left( \frac{b - a}{2} \right)\]
\[ = 2\cos\left( \frac{a + b}{2} \right)\sin\left( \frac{b - a}{2} \right)\]
\[ = \sin b - \sin a .....................\left[\text{Since, }2\cos A \sin B = \sin\left( A + B \right) - \sin\left( A - B \right) \right]\]