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(cos7x-cos8x)/(1+2cos5x).what is the integration of it?

Asked by harshita srivastava | 20th Jul, 2013, 10:37: PM

Expert Answer:

$\int {{{2\sin{{15x}\over 2}\sin{x\over 2}}\over {1+2-4\sin ^2 \frac {5x}2}}dx$

= $\int {{{2\sin {{15x}\over 2}\sin {x\over 2}}\over {3-4\sin ^2 \frac {5x}2}}dx$

= $\int {{{2\sin {{15x}\over 2}\sin {x\over 2}\sin {{5x}\over 2}}\over {3\sin \frac {5x}2}-4\sin ^3 \frac {5x}2}}dx$

= $\int {{{2\sin {{15x}\over 2}\sin {x\over 2}\sin {{5x}\over 2}}\over {\sin \frac {15x}2}}dx$

= $\int {2\sin {x\over 2}\sin {{5x}\over 2}dx$

= $\int {(\cos 2x-\cos 3x)dx$

= ${1\over 2}\sin 2x-{1\over 2}\sin 3x+C$

Answered by | 20th Jul, 2013, 10:59: PM

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