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Evaluation of $$\int \tan x\sqrt{1+\sin x}\mathrm dx$$

$\bf{My\; Try:}$

Let $(1+\sin x)= t^2\;$, then $$\displaystyle \cos x\,\mathrm dx= 2t\mathrm dt\implies\mathrm dx = \frac{2t}{\sqrt{2-t^2}}\mathrm dt$$

So, the integral becomes $$\displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}\mathrm dt = 2\int\frac{t^4-t^2}{2-t^2}\mathrm dt $$

Now, how can I proceed after this? Please help me.

Integreek
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juantheron
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6 Answers6

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Performing a polynomial long division followed by a partial fraction decomposition results in $$2\int\frac{t^4-t^2}{2-t^2}dt=-2\int(t^2+1)dt+\sqrt{2}\int\left(\frac{1}{\sqrt{2}-t}+\frac{1}{\sqrt{2}+t}\right)dt$$ I suppose you can take it from there.

Alijah Ahmed
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I think you made some mistakes is the substitution, $$t^2=\sin x +1$$ so $$2tdt=\cos x dx$$

now $$\frac{\tan x}{\cos x}= \frac{\sin x}{\cos^2 x}= \frac{\sin x}{1-\sin^2 x}= \frac{t^2-1}{1-(t^2-1)^2 }=\frac{t^2-1}{2t^2-t^4 }$$

So $$\int \tan x \sqrt{\sin x +1} dx= \int \frac{\tan x}{\cos x} \sqrt{\sin x +1} \cos x dx=\int \frac{t^2-1}{2t^2-t^4 } 2t^2dt= \int 2\frac{t^2-1}{2-t^2 } dt$$ To integrate $$\int \frac{t^2-1}{t^2-2}dt=\int 1+ \frac{1}{t^2-2}dt = \int 1+ \frac{1}{2\sqrt{2}} \left( \frac{1}{t-\sqrt{2}} -\frac{1}{t-\sqrt{2}} \right)dt $$

Now we can integrate getting some logarithms.

Rene Schipperus
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$$ 2\int\frac{t^4-t^2}{2-t^2}dt = 2\int\frac{\require{cancel}\cancel{(t^2 - 2)}(t^2 + 1)+2}{-(\cancel{t^2 - 2})},dt = \underbrace{-2\int (t^2 + 1)\,dt}_{\text{a cinch}} - \underbrace{2\int \frac {2}{t^2 - 2}\,dt}_{\text{partial fractions}}$$

Alternatively, for the second integral, we can express it as $$+ 2\cdot\frac{2}{1-t^2}$$ and use $t = \sin \theta \,d\theta \implies dt = \cos \theta \,d\theta $ to get $$4\int \sec \theta d\,\theta$$

amWhy
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Hint: $\frac{t^4-t^2}{2-t^2} \equiv -t^2-\frac{2}{t^2-2}-1,$ which is easy to integrate.

You should find, after a suitable hyperbolic substitution, that $\int \frac{2}{t^2-2}dt= \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2}-t}{\sqrt{2}+t} \right|$.

beep-boop
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Alternatively, integrate by parts \begin{align} \int \tan x &\sqrt{1+\sin x}\ dx =\int\sqrt{\frac{{1+\sin x}}{\cos x}}\ d(-2\sqrt{\cos x})\\ &=-2\sqrt{1+\sin x}+\int\frac{\sqrt{1+\sin x}}{\cos x} dx\\ &=-2\sqrt{1+\sin x}+\sqrt2\tanh^{-1}\frac{\sqrt{1+\sin x}}{\sqrt2} \end{align}

Quanto
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Alternatively, substitute $t=\sin x$:

$$\begin{align}\int\tan x\sqrt{1+\sin x}\,\mathrm dx&=\int\frac{t\sqrt{1+t}}{1-t^2}\mathrm dt\\&=\int\frac{t^2+t}{(1-t)\sqrt{1+t}}\mathrm dt\\&=-\int\sqrt{1+t}+\frac1{\sqrt{1+t}}\mathrm dt+2\int\frac{\mathrm dt}{(1-t)\sqrt{1+t}}\end{align}$$

The last integral can be evaluated by the substitution $u=\sqrt{1+t}$.

Integreek
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