Question

integral limit 0 to π/2( sinx - cosx /1+sinx-cosx ) dx

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Answer 1

EXPLANATION.

$\sf \implies \displaystyle I = \int\limits^{\pi/2}_{0} \bigg[ \frac{sin(x) - cos(x)}{1 + sin(x) - cos(x)} \bigg] dx . - - - - - (1).$

As we know that,

King Concept.

$\sf \implies \displaystyle \int\limits^b_a f(x)dx = \int\limits^b_a f(a + b - x) dx$

Using this concept in the equation, we get.

$\sf \implies \displaystyle I = \int\limits^{\pi/2}_{0} \bigg[\dfrac{sin(\pi/2 - x) - cos(\pi/2 - x)}{1 + sin(x) - cos(x)} \bigg] dx$

$\sf \implies \displaystyle I = \int\limits^{\pi/2}_{0} \bigg[ \frac{cos(x) - sin(x)}{1 + sin(x) - cos(x)} \bigg] dx. - - - - - (2).$

Adding equation (1) and (2), we get.

$\sf \implies \displaystyle 2I = \int\limits^{\pi/2}_{0} \bigg[ \frac{sin(x) - cos(x) + cos(x) - sin(x)}{1 + sin(x) - cos(x)} \bigg] dx$

$\sf \implies \displaystyle I = 0.$

$\sf \implies \displaystyle \boxed{\int\limits^{\pi/2}_{0} \bigg[ \frac{sin(x) - cos(x)}{1 + sin(x) - cos(x)} \bigg] dx = 0. }$

                                                                                                               

MORE INFORMATION.

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(sin\ x)dx = \int\limits^{\pi/2}_{0} ln(cos \ x)dx = \frac{- \pi}{2} ln(2).$

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(cosec\ x)dx = \int\limits^{\pi/2}_{0} ln(sec \ x)dx = \frac{\pi}{2} ln(2).$

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(tan \ x)dx = \int\limits^{\pi/2}_{0} ln(cot \ x)dx = 0.$