MvCalc10

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I=\int _{{0}}^{{{\frac  {\pi }{4}}}}[\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin(x+y)dx]dy\,

\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin(x+y)dx=[-\cos(x+y)]_{{0}}^{{{\frac  {\pi }{2}}}}\,

=[-\cos({\frac  {\pi }{2}}+y)-(-\cos y)]\,

=\sin y+\cos y\,

Now I=\int _{{0}}^{{{\frac  {\pi }{4}}}}(1+\sin y)dy\,

=[\sin y-\cos y]_{{0}}^{{{\frac  {\pi }{4}}}}\,

=[\sin {\frac  {\pi }{4}}-\cos {\frac  {\pi }{4}}]-[0-\cos 0]\,

={\frac  {1}{{\sqrt  {2}}}}-{\frac  {1}{{\sqrt  {2}}}}+1\,

=1\,

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