How to Evaluate Integral with Tan and Square Root?

[anemone]

Evaluate $\Large\int_0^{\pi/2}\frac{dx}{1+(\tan x)^{\sqrt{2}}}$.

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[anemone]

Re: Problem of the week #108 -April 21st, 2014

Congratulations to the following members for their correct solutions!:)

1. MarkFL
2. Opalg
3. kaliprasad
4. Pranav

Solution from MarkFL:

Spoiler

We are given to evaluate:

$$I=\int_0^{\pi/2}\frac{1}{1+\left(\tan(x)\right)^{\sqrt{2}}}\,dx$$

If we use the substitution:

$$u=\frac{\pi}{2}-x$$

and then use $x$ as the dummy variable instead, we obtain:

$$I=\int_0^{\pi/2}\frac{\left(\tan(x)\right)^{\sqrt{2}}}{1+\left(\tan(x)\right)^{\sqrt{2}}}\,dx$$

And so adding the two expressions, we obtain:

$$2I=\int_0^{\pi/2}\,dx=\frac{\pi}{2}$$

Hence:

$$I=\frac{\pi}{4}$$

Solution from kaliprasad:

Spoiler

We are given to evaluate:

$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2}x }\,dx$$

Using the property of definite integrals $$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$ and a co-function identity, we may state:

$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^\sqrt{2}(x)}\,dx$$

Adding the two equations, we obtain:

$$2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2} x }+\frac{1}{1+\cot^\sqrt{2} x}\,dx$$

$$2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^\sqrt{2} x+\cot^\sqrt{2} x}{2 + \tan^\sqrt{2} x+\cot^\sqrt{2} x}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}$$

Hence:

$$I=\frac{\pi}{4}$$