integration... what did i do wrong? (1 Viewer)

nat_doc · Jun 5, 2010

ok here is the question...
$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{cos^3x}{sin^2x}.dx_$
$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}cosx. cot^2x.dx$
$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}cosx.(sec^2x-1) .dx$
$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(secx-cosx) .dx$
$[ln(secx+tanx)-sinx]_{\pi/6}^{\pi/2}$

how how do i sub pi/2 in??

nat_doc · Jun 5, 2010

Handmedown · Jun 5, 2010

nat_doc · Jun 5, 2010

Handmedown said:
$1 + cot^{2}x = cosec^{2}x$

freaking silly me..

i love you!.

lolman12567 · Jun 5, 2010

you could also change the integrand into cosx.(1-(sinx)^2)/(sinx)^2 then let u=sinx

integration... what did i do wrong? (1 Viewer)

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