Integral Calculus> Integrate ∫sqroot(tanx) dx Reply soon.......

Integrate
∫sqroot(tanx) dx
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Lovey , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Ans:Hello student, please find answer to your question
$I = \int \sqrt{tanx}.dx$
$tanx = t^{2}$
$sec^{2}x.dx = 2t.dt$
$(t^{4}+1).dx = 2t.dt$
$I = \int \frac{2t^{2}}{t^{4}+1}dt$
$I = \int \frac{t^{2}-1+t^{2}+1}{t^{4}+1}dt$
$I = \int \frac{t^{2}-1}{t^{4}+1}dt+\int \frac{t^{2}+1}{t^{4}+1}dt$
$I = \int \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}dt+\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}dt$
$I = \int \frac{1-\frac{1}{t^{2}}}{(t+\frac{1}{t})^{2}-(\sqrt{2})^{2}}dt+\int \frac{1+\frac{1}{t^{2}}}{(t-\frac{1}{t})^{2}+(\sqrt{2})^{2}}dt$
$I = \frac{1}{2\sqrt{2}}log|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}|+\frac{1}{\sqrt{2}}tan^{-1}(\frac{t-\frac{1}{t}}{\sqrt{2}}) + constant$
$I = \frac{1}{2\sqrt{2}}log|\frac{\sqrt{tanx}+\sqrt{cotx}-\sqrt{2}}{\sqrt{tanx}+\sqrt{cotx}+\sqrt{2}}|+\frac{1}{\sqrt{2}}tan^{-1}(\frac{\sqrt{tanx}-\sqrt{cotx}}{\sqrt{2}}) + constant$