# 11 ) integ [tan 2θ / √cos6θ + sin6θ  ] dθ12 ) integ [ cos2x / (1 + tanx) ] dx13 ) integ [ { ln (ln (1 + x)/(1 - x) ) ) } / ( 1 - x2 ) ] dx14 ) integ { [ (x/e)x + (e/x)x ] lnx } dx

Jitender Singh IIT Delhi
9 years ago
Ans:
$I = \int \frac{cos^{2}x}{1+tanx}dx$
$I = \int \frac{cos^{2}x.sec^{4}x}{(1+tanx).sec^{4}x}dx$
$I = \int \frac{sec^{2}x}{(sec^{4}x+sec^{4}x.tanx)}dx$
$I = \int \frac{sec^{2}x}{(1+tanx)(1+tan^{2}x)^{2}}dx$
$tanx = t$
$sec^{2}x.dx = dt$
$I = \int \frac{1}{(1+t)(1+t^{2})^{2}}dt$
$I = \int (\frac{1-t}{4.(1+t^{2})}+\frac{1-t}{2.(1+t^{2})^{2}}+\frac{1}{4.(t+1)})dt$
$I = \frac{(t^{2}+1).(-log(t^{2}+1))+2log(t+1)+4tan^{-1}t+sin(2tan^{-1}t)+2}{8(t^{2}+1)}$$I = \frac{1}{8}(4x+sin2x+cos2x+2log(sinx+cosx))+constant$
$I = \int \frac{tan2\theta }{\sqrt{cos^{6}\theta +sin^{6}\theta} }d\theta$
$I = tanh^{-1}(\frac{\sqrt{3cos4\theta +5}}{\sqrt{2}}) + constant$
$I = \int \frac{log(log(\frac{1+x}{1-x}))}{1-x^{2}}.dx$
$log(\frac{1+x}{1-x}) = t$
$I = \frac{1}{2}\int log(t) . dt$
$I = \frac{1}{2}t.log(t) - \frac{t}{2}+constant$
$I = \frac{1}{2}log(\frac{1+x}{1-x})(log(log(\frac{1+x}{1-x})-1))+constant$
There is some mistakes in last integrand.
Thanks & Regards
Jitender Singh
IIT Delhi