G
Guest
Guest
Hi
I need to evaluate this definite integral:
\(\displaystyle \L\int_0^{{\textstyle{\pi \over 2}}} {\cos 3x\sin 5xdx}\)
This is what I got
\(\displaystyle \L\int_0^{{\textstyle{\pi \over 2}}} {\cos 3x\sin 5xdx} = \int_0^{{\textstyle{\pi \over 2}}} {\sin 5x\cos 3xdx}\)
\(\displaystyle \L = \int_0^{{\textstyle{\pi \over 2}}} {{\textstyle{1 \over 2}}\left\{ {\sin (5x + 3x) + \sin (5x - 3x)} \right\}dx}\)
\(\displaystyle \L = {\textstyle{1 \over 2}}\int_0^{{\textstyle{\pi \over 2}}} {\left\{ {\sin 8x + \sin 2x)} \right\}} dx\)
\(\displaystyle \L = \left[ { - {\textstyle{1 \over {16}}}\cos 8x - {\textstyle{1 \over 4}}\cos 2x} \right]_0^{{\textstyle{\pi \over 2}}}\)
\(\displaystyle \L = - {\textstyle{1 \over {16}}}\left[ {\cos 8x + 4\cos 2x} \right]_0^{{\textstyle{\pi \over 2}}}\)
\(\displaystyle \L
= - {\textstyle{1 \over {16}}}\left[ {\left\{ {\cos (8)({\textstyle{\pi \over 2}}) - \cos (8)(0)} \right\} + 4\left\{ {\cos (2)({\textstyle{\pi \over 2}}) - \cos (2)(0)} \right\}} \right]\)
\(\displaystyle \L\begin
= - {\textstyle{1 \over {16}}}\left[ {(1 - 1) + 4( - 1 - 1)} \right] = - {\textstyle{1 \over {16}}}\left[ { - 8} \right] \\
= \frac{1}{2} \\\)
Thanks for checking it for me
I need to evaluate this definite integral:
\(\displaystyle \L\int_0^{{\textstyle{\pi \over 2}}} {\cos 3x\sin 5xdx}\)
This is what I got
\(\displaystyle \L\int_0^{{\textstyle{\pi \over 2}}} {\cos 3x\sin 5xdx} = \int_0^{{\textstyle{\pi \over 2}}} {\sin 5x\cos 3xdx}\)
\(\displaystyle \L = \int_0^{{\textstyle{\pi \over 2}}} {{\textstyle{1 \over 2}}\left\{ {\sin (5x + 3x) + \sin (5x - 3x)} \right\}dx}\)
\(\displaystyle \L = {\textstyle{1 \over 2}}\int_0^{{\textstyle{\pi \over 2}}} {\left\{ {\sin 8x + \sin 2x)} \right\}} dx\)
\(\displaystyle \L = \left[ { - {\textstyle{1 \over {16}}}\cos 8x - {\textstyle{1 \over 4}}\cos 2x} \right]_0^{{\textstyle{\pi \over 2}}}\)
\(\displaystyle \L = - {\textstyle{1 \over {16}}}\left[ {\cos 8x + 4\cos 2x} \right]_0^{{\textstyle{\pi \over 2}}}\)
\(\displaystyle \L
= - {\textstyle{1 \over {16}}}\left[ {\left\{ {\cos (8)({\textstyle{\pi \over 2}}) - \cos (8)(0)} \right\} + 4\left\{ {\cos (2)({\textstyle{\pi \over 2}}) - \cos (2)(0)} \right\}} \right]\)
\(\displaystyle \L\begin
= - {\textstyle{1 \over {16}}}\left[ {(1 - 1) + 4( - 1 - 1)} \right] = - {\textstyle{1 \over {16}}}\left[ { - 8} \right] \\
= \frac{1}{2} \\\)
Thanks for checking it for me