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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\frac{\Large \pi}{\Large 2}\frac{\sin x-\cos x}{1+\sin x\cos x}\;dx\]

$\begin{array}{1 1}\large \frac{\pi}{2} \\ \large \frac{\pi}{4} \\ \pi \\ 0 \end{array} $

1 Answer

Toolbox:
  • (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin x-\cos x}{1+\sin x\cos x}\;dx-----(1)$
 
Applying the property $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
 
$I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin (\pi/2-x)-\cos (\pi/2-x)}{1+\sin (\pi/2-x)\cos (\pi/2-x)}\;dx$
 
But $\sin (\pi/2-x)=\cos x\;and\;\cos(\pi/2-x)=\sin x$
 
Therefore $I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\cos x-\sin x}{1+\cos x\sin x}\;dx-----(2)$
 
Adding equ(1) and equ(2)
 
$2I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin x-\cos x}{1+\sin x\cos x}+\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\cos x-\sin x}{1+\cos x\sin x}$
 
On simplifying,
 
$2I=0\qquad=>I=0$

 

 

answered Feb 14, 2013 by meena.p
edited Feb 14, 2013 by meena.p
 

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