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Evaluate the following:$\int\limits_0^{\Large\pi} \large \frac{x}{1+\sin x}$

1 Answer

  • $\int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
  • $\int \sec^2 x dx=\tan x+c$
  • $\int \sec x \tan x dx=\sec x+c$
Step 1:
Let $I=\int\limits_0^{\Large\pi} \large\frac{x}{1+\sin x}$-----(1)
By applying the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x) dx$
we get
$I=\int\limits_0^{\Large\pi} \large\frac{\pi-x}{1+\sin (\pi-x)}$-----(2)
But $\sin (\pi-x) =\sin x$
Now adding equ (1) and equ(2)
$2I=\int\limits_0^{\Large\pi} \large\frac{x+\pi-x}{1+\sin x}$$dx$
$=\pi \int\limits_0^{\Large\pi} \large\frac{dx}{1+\sin x}$
Mutiply and divide by $1-\sin x$
$2I=\pi \int\limits_0^{\Large\pi} \large\frac{(1-\sin x)dx}{1-\sin ^2 x}$
$=\pi \int\limits_0^{\Large\pi} \large\frac{(1-\sin x)dx}{\cos ^2 x}$
Now splitting the terms
$2I=\pi \bigg[ \int\limits_0^{\Large\pi} \large\frac{1}{\cos ^2 x}$$dx- \int\limits_0^{\Large\pi} \large\frac{\sin x}{\cos ^2 x}$$dx\bigg]$
But $\large\frac{1}{\cos ^2 x}=$$\sec^2 x$ and
$\large\frac{\sin x}{\cos ^2 x}=$$\tan x.\sec x$
$2I=\pi \bigg[ \int\limits_0^{\Large\pi} \sec ^2 x $$dx- \int\limits_0^{\Large\pi} \sec x \tan x dx\bigg]$
Step 2:
On integrating we get
$2I= \pi \bigg\{ \bigg[\tan x\bigg]_0^{\pi}-\bigg[\sec x\bigg]_0^{\pi} \bigg\}$
On Applying limits we get,
$2I= \pi \bigg\{ \bigg[\tan \pi-\tan 0\bigg]-\bigg[\sec \pi-\sec 0\bigg] \bigg\}$
$\tan \pi=\tan 0\;and\; sec \pi=-1\; and \;sec 0=1$
Hence $2I=\pi \{0-[-1-1]\}$
$2I=2 \pi \qquad => I=\pi$


answered Apr 20, 2013 by meena.p