Evaluate the integral ∫0π/211+tanα⁡xdx

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MoxboasteBots5h · Accepted Answer

The integral can be solved by using the fact that tan⁡(π2−x)=1tan⁡x∫abf(x)dx=∫abf(a+b−x)dxLetI(α)=∫0π2dx1+tanαxthenI(α)=∫0π2dx1+tanα(π2+0−x)=∫0π2dx1+1tanαx=∫0π2tanαx1+tanαxdxAdding the two I(α)′s yields2I(α)=∫0π211+tanαxdx+∫0π2tanαx1+tanαxdx=∫0π2dx=π2I(α)=π4

Mason Hall · Accepted Answer

Set I(α)=∫0π2f(α,x)dx where f(α,x):R×(0,π2)→R is defined as f(α,x)=11+tanα(x)Since both ddαf(α,x)=−tanα(x)log⁡(tan⁡(x))(1+tanα(x))2 and f(α,x) are continuous on (0,π2)→R we can apply the Leibniz integral rule obtainingI′(α)=ddα∫0π2f(α,x)dx=∫0π2ddαf(α,x)dx=0since ddαf(α,π4+x)=ddαf(α,π4−x) for x∈[0,π4).Jence I(α) is constant on R andI(α)=I(0)=∫0π2dx1+1=π4

Vasquez · Accepted Answer

Well, this problem has a nice symmetry. The integral can be rewritten as,
$I = \int_{0}^{π / 2} \frac{\cos^{α} (x)}{\sin^{α} (x) + \cos^{α} (x)} d x$
By the property of definite integrals this Integral is same as,
$\int_{0}^{π / 2} \frac{\cos^{α} (\frac{π}{2} - x)}{\sin^{α} (\frac{π}{2} - x) + \cos^{α} (\frac{π}{2} - x)} d x$
$\int_{0}^{π / 2} \frac{\sin^{α} (x)}{\sin^{α} (x) + \cos^{α} (x)}$
adding to equation we get
$2 I = \int_{0}^{π / 2} d x$
$2 I = \frac{π}{2}$
$I = \frac{π}{4}$

Evaluate the integral\int_0^{\pi/2}\frac{1}{1+\tan^\alfa x}dx

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