Q: The following integral \$\displaystyle \int_{\pi/4}^{\pi/2} (2 cosec x)^{17}dx\$ is equal to

\$\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 cosec x)^{17}dx\$Let\$e^u+e^{-u}= 2 cosec x\$\$ x = \displaystyle \frac{\pi}{4} \Rightarrow u = ln (1+\sqrt{2})\$\$x = \dfrac{\pi}{2} \Rightarrow u = 0\$\$\Rightarrow cosec x + cot x = e^u\$ and \$cosec x - cot x = e^{-u} \$\$\Rightarrow x = \displaystyle \frac{e^u - e^{-u}}{2} (e^u - e^{-u}) dx = -2 cosec x cot x dx\$\$\Rightarrow - \displaystyle \int(e^u + e^{-u})^{17} \frac{(e^u - e^{-u})}{2 cosec x cot x}du\$\$=-2 \int_{ln(1+\sqrt{2})}^{0} (e^u + e^{-u})^{16}du\$\$= \int_{0}^{ln (1+\sqrt{2})} 2(e^u + e^{-u})^{16}du\$

Q: The value of \$g\displaystyle \left ( \frac{1}{2} \right )\$ is?

\$g \displaystyle \left ( \frac{1}{2} \right ) = \lim_{h \rightarrow 0^+} \int_h^{1 - h} t^{-1/2} (1-t)^{-1/2} dt\$\$= \int_0^1 \displaystyle \frac{dt}{\sqrt{t - t^2}} = \int_0^1 \frac{dt}{\sqrt{\frac{1}{4} - \left ( t-\frac{1}{2} \right )^2}} = sin^{-1} \left ( \frac{t - \frac{1}{2}}{\frac{1}{2}} \right ) |_0^1\$\$= sin^{-1} 1 -sin^{-1} (-1) = \pi\$

Question 1

The following integral $\displaystyle \int_{\pi/4}^{\pi/2} (2 cosec  x)^{17}dx$ is equal to

Accepted Answer

$\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 cosec  x)^{17}dx$Let$e^u+e^{-u}= 2 cosec  x$$ x = \displaystyle \frac{\pi}{4}  \Rightarrow u = ln (1+\sqrt{2})$$x = \dfrac{\pi}{2}  \Rightarrow u = 0$$\Rightarrow cosec  x + cot  x = e^u$  and  $cosec  x - cot  x = e^{-u} $$\Rightarrow  x = \displaystyle \frac{e^u - e^{-u}}{2} (e^u - e^{-u}) dx = -2  cosec x cot  x dx$$\Rightarrow - \displaystyle \int(e^u + e^{-u})^{17} \frac{(e^u - e^{-u})}{2  cosec  x  cot  x}du$$=-2  \int_{ln(1+\sqrt{2})}^{0}  (e^u + e^{-u})^{16}du$$= \int_{0}^{ln (1+\sqrt{2})} 2(e^u + e^{-u})^{16}du$

Question 2

The value of $g\displaystyle \left ( \frac{1}{2} \right )$ is?

Accepted Answer

$g \displaystyle \left ( \frac{1}{2} \right ) = \lim_{h \rightarrow 0^+} \int_h^{1 - h} t^{-1/2} (1-t)^{-1/2} dt$$= \int_0^1 \displaystyle \frac{dt}{\sqrt{t - t^2}} = \int_0^1 \frac{dt}{\sqrt{\frac{1}{4} - \left ( t-\frac{1}{2} \right )^2}} = sin^{-1} \left ( \frac{t - \frac{1}{2}}{\frac{1}{2}} \right ) |_0^1$$= sin^{-1} 1 -sin^{-1} (-1) = \pi$

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