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11 ) integ [tan 2θ / √ cos 6 θ + sin 6 θ ] dθ 12 ) integ [ cos 2 x / (1 + tanx) ] dx 13 ) integ [ { ln (ln (1 + x)/(1 - x) ) ) } / ( 1 - x 2 ) ] dx 14 ) integ { [ (x/e) x + (e/x) x ] lnx } dx

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

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 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
askIITians Faculty

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