`int_0^(pi/2) cos(t)/(sqrt(1 + sin^2(t))) dt` Evaluate the integral

Saturday, February 23, 2013

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{\cos{{\left({t}\right)}}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{\left({t}\right)}}}}}{\left.{d}{t}\right.}$ Evaluate the integral

You need to perform the following substitution to solve the integral $\displaystyle{\sin{{t}}}={u}\Rightarrow{\cos{{t}}}{\left.{d}{t}\right.}={d}{u}\Rightarrow{t}={\arcsin{{u}}}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{d}{u}}}{{\sqrt{{{1}+{{u}}^{{2}}}}={\ln{{\left({u}+\sqrt{{{{u}}^{{2}}+{1}}}\right)}}}{{\mid}_{{{u}_{{1}}}}^{{{u}_{{2}}}}}}}$

Replacing back u for t yields:

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({\sin{{t}}}+\sqrt{{{1}+{{\sin}}^{{2}}{t}}}\right)}}}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({\sin{{\left(\frac{\pi}{{2}}\right)}}}+\sqrt{{{1}+{{\sin}}^{{2}}{\left(\frac{\pi}{{2}}\right)}}}\right)}}}-{\ln{{\left({\sin{{\left({0}\right)}}}+\sqrt{{{1}+{{\sin}}^{{2}}{0}}}\right)}}}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({1}+\sqrt{{2}}\right)}}}-{\ln{{\left({0}+{1}\right)}}}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({1}+\sqrt{{2}}\right)}}}-{\ln{{1}}}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({1}+\sqrt{{2}}\right)}}}-{0}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({1}+\sqrt{{2}}\right)}}}$

Hence, evaluating the definite integral yields $\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\cos{{t}}}{\left.{d}{t}\right.}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{t}}}}}={\ln{{\left({1}+\sqrt{{2}}\right)}}}.$

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Saturday, February 23, 2013

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{\cos{{\left({t}\right)}}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{\left({t}\right)}}}}}{\left.{d}{t}\right.}$ Evaluate the integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Saturday, February 23, 2013

Evaluate the integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{\cos{{\left({t}\right)}}}}{{\sqrt{{{1}+{{\sin}}^{{2}}{\left({t}\right)}}}}}{\left.{d}{t}\right.}$ Evaluate the integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?