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I have trouble calculating the following definite integral $$\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1 + \sqrt{\tan x}} \text{d}x.$$ I tried to get the primitive function by $t = \tan \dfrac{x}{2}$ but it seemed that it is too complicated.

Qetuo_233
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  • https://www.wolframalpha.com/input?i=integrate+sinxcosx%2F%281%2Bsqrt%28tanx%29%29 there is an elementary primitive though! the defininite evaluates to 1/4 using this primtive, but i'm sure there are smarter methods given how tedious this would be – Max0815 Jan 06 '23 at 07:04
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    As a hint (and this a pretty useful trick): show for any $a,b$ and function $f$, $\int_a^b f(x) \mathrm{d}x = \int_a^b f(a+b-x) \mathrm{d}x.$ Call the integral $I$, and apply this to get a second expression for $I$. Can you use these two expressions together to get something tractable? – stochasticboy321 Jan 06 '23 at 07:20
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    See https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Jan 06 '23 at 07:42
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    Try $\sqrt{\tan(x)}=t$ for the antiderivative – Claude Leibovici Jan 06 '23 at 08:56
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    Integrate[Sin[x]*Cos[x]/(1 + Sqrt[Tan[x]]), {x, 0, Pi/2}] $\frac{1}{4}$ – 138 Aspen Jan 06 '23 at 11:25
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    $$I_0=\int_0^{\pi / 2} \frac{\sin x \cos x}{1 + \sqrt{\tan x}}dx=\int_0^{\pi / 2} \frac{\tan x \cos^2 x}{1 + \sqrt{\tan x}}dx\overset{t=\tan x}{=}\int_0^\infty\frac{t}{(1+t^2)^2}\frac{dt}{1+\sqrt t}$$ $$\overset{IBP}{=}-\frac{1}{2}\frac{1}{1+t^2}\frac{1}{1+\sqrt t},\bigg|_0^\infty-\frac{1}{2}\int_0^\infty\frac{1}{1+x^4}\frac{dx}{(1+x)^2}$$ $$I_1=\int_0^\infty\frac{1}{1+x^4}\frac{dx}{(1+x)^2}\overset{x=\frac{1}{t}}{=}\int_0^\infty\frac{t^4}{1+t^4}\frac{dt}{(1+t)^2}\Rightarrow 2I_1=\int_0^\infty\frac{1+t^4}{1+t^4}\frac{dt}{(1+t)^2}=1$$ $$I_0=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$ – Svyatoslav Jan 06 '23 at 12:43

2 Answers2

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Expanding on @stochasticboy321's comment, you get $$ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)\cos(x)}{1+\sqrt{\tan(x)}}\mathrm{d}x + \underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\cos(u)\sin(u)}{1+\sqrt{\cot(u)}}\mathrm{d}u}_{\color{darkblue}{u = \frac{\pi}{2}-x}} = \int_{0}^{\frac{\pi}{2}} \sin(x)\cos(x) \mathrm{d}x = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \sin(2x) \mathrm{d}x $$ Since $$ \frac{1}{1+\sqrt{t}} + \frac{1}{1+\sqrt{\frac{1}{t}}} = \frac{1}{1+\sqrt{t}} + \frac{\sqrt{t}}{\sqrt{t}+1} = \frac{1+\sqrt{t}}{1+\sqrt{t}}=1 $$

Robert Lee
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Let $\tan x=u^2$ then $$I=\int_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sqrt{\tan x}}=\int_0^{\infty}\frac{2u^3}{(u+1)(u^4+1)^2}du$$ We can still do a similar trick. It is not too late. Let $u\rightarrow 1/u$ in the last integral, then, $$I=\int_0^{\infty}\frac{2u^4}{(u+1)(u^4+1)^2}du$$ and hence $$I=\frac{1}{2}\int_0^{\infty}\frac{2u^3+2u^4}{(u+1)(u^4+1)^2}du=\int_0^{\infty}\frac{u^3}{(u^4+1)^2}du=\left.-\frac{1}{4(u^4+1)}\right\rvert_0^{\infty}=\frac{1}{4}.$$

Bob Dobbs
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