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

Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$

$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$

So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}dt = 2\int\frac{t^4-t^2}{2-t^2}dt $

Now How Can I solve after that

Help me

Thanks

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $\int \frac{2}{t^2-2}dt= \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2}-t}{\sqrt{2}+t} \right|$.

Solution 4:

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

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 $$4\int \sec \theta d\,\theta$$