MCQ
$\int_0^{\pi /2} {\frac{{\sin x}}{{\sin x + \cos x}}\,dx} $ equals
- A$\frac{\pi }{2}$
- B$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
- D$\frac{\pi }{6}$
$\,\,\left( \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right)$
$2I = \int_0^{\pi /2} {dx} \Rightarrow I = \frac{\pi }{4}$.
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| $X = x_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | |
| $P(X = X_i)$ | $0$ | $2p$ | $2p$ | $3p$ | $p^2$ | $2p^2$ | $7p^2$ | $7p^2$ | $2p$ |
(where $C$ is a constant of integration)
$\int\limits_{ - \,1}^x {\,\left( {8{t^2} + \frac{{28}}{3}t + 4} \right)\,dt} $ $=$ $\frac{{\left( {{\textstyle{3 \over 2}}} \right)x + 1}}{{{{\log }_{(x + 1)}}\sqrt {x + 1} }}$ , is