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Question Number 136813 by mnjuly1970 last updated on 26/Mar/21

\dots \dots a d v a n c e d c a l c u l u s (I) \dots .

i f :

Ω = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{x . c o s (2 x) . c o s (x)}{{s i n}^{7} (x)} d x = \frac{a π}{90}

t h e n :: a = ? ? ?

Answered by Ar Brandon last updated on 26/Mar/21

Ω = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{xcos 2 xcosx}{\sin^{7} x} dx = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{x (1 - {2 \sin}^{2} x) cosx}{\sin^{7} x} dx

= \int_{\frac{π}{4}}^{\frac{π}{2}} {x \cdot \frac{cosx}{\sin^{7} x} - x \cdot \frac{2 cosx}{\sin^{5} x}} dx

= {[x \cdot \frac{1}{- {6 \sin}^{6} x} + \frac{1}{6} \int \frac{dx}{\sin^{6} x}] - 2 [x \cdot \frac{1}{- {4 \sin}^{4} x} + \frac{1}{4} \int \frac{dx}{\sin^{4} x}]}_{\frac{π}{4}}^{\frac{π}{2}}

= - \frac{π}{2} \cdot \frac{1}{6} + \frac{π}{4} \cdot \frac{8}{6} + 2 (\frac{π}{2} \cdot \frac{1}{4} - \frac{π}{4}) + \frac{1}{6} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{dx}{\sin^{6} x} - \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{dx}{\sin^{4} x}

= \frac{1}{6} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{dx}{\sin^{6} x} - \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{dx}{\sin^{4} x}

= \frac{1}{6} \int_{\frac{π}{4}}^{\frac{π}{2}} {cosec}^{2} x {(1 + \cot^{2} x)}^{2} dx - \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} ({cosec}^{2} x) (\cot^{2} x + 1) dx

= {[\frac{1}{2} (\frac{\cot^{3} x}{3} + cotx) - \frac{1}{6} (cotx + 2 \frac{\cot^{3} x}{3} + \frac{\cot^{5} x}{5})]}_{\frac{π}{4}}^{\frac{π}{2}} = - \frac{16}{45} = \frac{a π}{90}, a = - \frac{32}{π}

Commented by mnjuly1970 last updated on 26/Mar/21

t h a n k s a l o t \dots

Commented by Ar Brandon last updated on 26/Mar/21

You're welcome Sir !