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

1 Answer

  • (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
Given $I=\int\limits_0^\pi\frac{x\;dx}{1+\sin x}-----(1)$
Applying the property $\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
But $\sin(\pi -x)=\sin x$
$I=\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx-----(2)$
Adding equ (1) and equ(2)
$2I=\int\limits_0^\pi\frac{xdx}{1+\sin x}+\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx$
On simplifying we get,
$2I=\int\limits_0^\pi\frac{\pi}{1+\sin x}dx$
Multiply and divide by $(1-\sin x)$
$2I=\pi\int\limits_0^\pi\frac{(1-\sin x)}{1+\sin x)(1-\sin x)}dx$
$(1+\sin x)(1-\sin x)=1-\sin^2x=\cos ^2x$
Therefore $2I=\pi\int\limits_0^\pi\frac{1-\sin x}{\cos^2 x}$
On seperating the terms,
$2I=\pi\int\limits_0^\pi\large\frac{1}{\cos^2 x}-\int\limits_0^\pi\frac{\sin x}{\cos x}.\frac{1}{\cos x}dx$
$=\pi\int\limits_0^\pi \sec ^2x-\int \limits_0^\pi \sec x.\tan x$
On integrating we get
$2I=\pi\bigg\{[\tan x]_0^\pi -[\sec x]_0^\pi \bigg\}$
On applying limits,
$2I=\pi[(\tan \pi-\tan 0)-(\sec \pi-\sec 0)]$
But $\tan \pi=\tan 0=0$
and $\sec \pi=-1\;and\; \sec 0=1$
Now substituting these values,
Therefore $I=\pi$



answered Feb 14, 2013 by meena.p

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