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Integrate 3 sec^2x log(tanx) dx.(a) -log(tanx) (tanx-1)+C(b) log(tanx) (secx+1)+C(c) tanx (log(tanx)-1)+C(d) tanx (logsecx +1)+CI had been asked this question in my homework.My question is based upon Integration by Parts topic in division Integrals of Mathematics – Class 12 |
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Answer» RIGHT answer is (C) tanx (log(tanx)-1)+C Easiest EXPLANATION: By using∫ U.v dx=u∫ v dx-∫ u'(∫ v dx), we get ∫ log(tanx) sec^2x dx=log(tanx) ∫ sec^2 xdx -∫ (logtanx)’∫ sec^2x dx =tanx log(tanx)-\(\int \frac{1}{tanx} sec^2x.tanx \,dx\) =tan xlog(tanx)-∫ sec^2x dx =tan xlog(tanx)-tanx+C =tanx (log(tanx)-1)+C |
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