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Integral Calculus> INTEGRATION SIKHO...

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11 ) integ [tan 2θ / √cos⁶θ + sin⁶θ ] dθ

12 ) integ [ cos²x / (1 + tanx) ] dx

13 ) integ [ { ln (ln (1 + x)/(1 - x) ) ) } / ( 1 - x² ) ] dx

14 ) integ { [ (x/e)^x + (e/x)^x ] lnx } dx

UTTARA P , 15 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 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

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