Mathemerize

Solution : Let I = \(\int_{0}^{\pi/2}\) log(sinx)dx    …….(i) then I = \(\int_{0}^{\pi/2}\) \(log(sin({\pi\over 2}-x))\)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx     …….(ii) Adding (i) and (ii), we get 2I = \(\int_{0}^{\pi/2}\) log(sinx)dx + \(\int_{0}^{\pi/2}\) log(cosx)dx = \(\int_{0}^{\pi/2}\) (log(sinx)dx + log(cosx)) \(\implies\) \(\int_{0}^{\pi/2}\) log(sinxcosx)dx = \(\int_{0}^{\pi/2}\) \(log({2sinxcosx\over 2})\)dx = \(\int_{0}^{\pi/2}\) \(log({sin2x\over 2})\)dx = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\int_{0}^{\pi/2}\) log(2)dx …

Prove that \(\int_{0}^{\pi/2}\) log(sinx)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx = -\(\pi\over 2\)log 2. Read More »

Solution : I = \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\) = \(\int\) \(cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}\) = \(\int\) \(cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}\) Put \(1+cot^5x\) = t \(5cot^4xcosec^2x\)dx = -dt = -\(1\over 5\) \(\int\) \(dt\over {t^{3/5}}\) = -\(1\over 2\) \(t^{2/5}\) + C = -\(1\over 2\) \({(1+cot^5x)}^{2/5}\) + C Similar Questions What is the integration …

Evaluate : \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\) Read More »

Solution : I = \(\int\) \(dx\over {3sinx + 4cosx}\) = \(\int\) \(dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}\) = \(\int\) \(sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}\) let \(tan{x\over 2}\) = t, \(\therefore\)  \({1\over 2}sec^2{x\over 2}\)dx = dt so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\) …

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\) Read More »