How to find∫π2π4cot5xcsc3xdx

Question

Donald Erickson · Accepted Answer

Note that it’s not necessary to convert to sines and cosines:

\cot^{5} x \csc^{3} x = \cot^{4} x \csc^{2} x (\cot x \csc x)

= {(\csc^{2} x - 1)}^{2} \csc^{2} x (\csc x \cot x)

and

d (\csc x) = - \csc x \cot x d x

, so you can simply let

u = \csc x

. The indefinite integral then becomes

\int {(\csc^{2} x - 1)}^{2} \csc^{2} x (\csc x \cot x) d x = - \int {(u^{2} - 1)}^{2} u^{2} d u

= - \int (u^{6} - 2 u^{4} + u^{2}) d u

= - \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3} + C

= - \frac{1}{7} \csc^{7} x + \frac{2}{5} \csc^{5} x - \frac{1}{3} \csc^{3} x + C

razlikaml42 · Accepted Answer

Here&#039;s one way to find the indefinite integral.Write everything in terms of sin⁡ and cos⁡. If one of those functions is raised to an odd positive power, say cos⁡ out and use the Pythagorean identity to write the other factor in terms of sin⁡. Then use a u-substitution with u=sin⁡x:∫cot5xcsc3xdx=∫cos5xsin5x1sin3xdx=∫cos⁡x⋅(cos2x)2sin8xdx=∫cos⁡x⋅(1−sin2x)2sin8xdx=∫(1−u2)2u8du=∫1−2u2+u4u8du=∫(u−8−2u−6+u−4)du=−u77+2u−55−u−33+C=−sin−7x7+2sin−5x5−sin−33+C

How to find \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\cot^5 x\csc^3 xdx

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Answer & Explanation

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