INTEGRAL

thaku sir [1]

Thnk u sir. I needed it.

We have
I=\int_0^\pi\dfrac{x}{1+\sin x\cos x}\ \mathrm dx = I_1+I_2
where
I_1=\int_0^{\pi/2}\dfrac{x}{1+\sin x\cos x}\ \mathrm dx
and
I_2=\int_{\pi/2}^\pi\dfrac{x}{1+\sin x\cos x}\ \mathrm dx
Consider I₁. Applying a property, we have
I_1=\int_0^{\pi/2}\dfrac{x}{1+\sin x\cos x}\ \mathrm dx =\int_0^{\pi/2}\dfrac{\frac{\pi}{2}-x}{1+\sin x\cos x}\ \mathrm dx
Hence,
2I_1=\dfrac{\pi}{2}\int_0^{\pi/2}\dfrac{1}{1+\sin x\cos x}\ \mathrm dx
So that
I_1=\dfrac{\pi}{4}\int_0^{\pi/2}\dfrac{1}{1+\sin x\cos x}\ \mathrm dx =\dfrac{\pi}{4}J_1
where
J_1=\int_0^{\pi/2}\dfrac{1}{1+\sin x\cos x}\ \mathrm dx =\int_0^{\pi/2}\dfrac{\sec^2x}{1+\tan x+\tan^2x}\ \mathrm dx=\int_0^\infty \dfrac{\mathrm dt}{1+t+t^2}
where the substitution tan x = t has been used.
Complete the square as
t^2+t+1=\left(t+\frac{1}{2}\right)^2+\dfrac{3}{4}
So that
J_1=\int_0^\infty \dfrac{\mathrm dt}{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\dfrac{2}{\sqrt{3}}\left|\tan^{-1}\dfrac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|_0^\infty=\dfrac{2}{\sqrt{3}}\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)
=\dfrac{4\pi}{6\sqrt{3}}
Hence, we get
\boxed{I_1=\dfrac{\pi^2}{6\sqrt{3}}}
Next, for I₂, first apply substitution
x=y+\frac{\pi}{2}
which gives \mathrm dx=\mathrm dy. Also when x=\pi/2, then y=0 and when x=\pi, then y=\pi/2
Hence, we get
I_2=\int_0^{\pi/2}\dfrac{\frac{\pi}{2}+y}{1-\sin y\cos y}\ \mathrm dy
Apply a property, we get
I_2=\int_0^{\pi/2}\dfrac{\frac{\pi}{2}+\frac{\pi}{2}-y}{1-\sin y\cos y}\ \mathrm dy
Hence,
2I_2=\dfrac{3\pi}{2}\int_0^{\pi/2}\dfrac{1}{1-\sin y\cos y}\ \mathrm dy which gives
I_2=\dfrac{3\pi}{4}\int_0^{\pi/2}\dfrac{1}{1-\sin y\cos y}\ \mathrm dy
This could again be integrated by the substitution tan y = z. The result is
\boxed{I_2=\dfrac{6\pi^2}{6\sqrt{3}}}
Hence, the given integral
\boxed{I=I_1+I_2=\dfrac{7\pi^2}{6\sqrt{3}}}

From UTTARA's method,
I_{1}=\int_{0}^{\pi }{\frac{-x(2sinxcosx)}{1- sin^{2}xcos^{2}x}}dx

Using f(x) = f(a+b-x) where a,b r limits of integral

I_{1}=\int_{0}^{\pi }{\frac{-(\pi -x)(2sin(\pi -x)cos(\pi -x))}{1- sin^{2}(\pi -x)cos^{2}(\pi -x)}}dx

I_{1}=\int_{0}^{\pi }{\frac{(\pi -x)(2sinxcosx)}{1- sin^{2}xcos^{2}x}}dx

I_{1}=\int_{0}^{\pi }{\frac{\pi(2sinxcosx)}{1- sin^{2}xcos^{2}x}}dx - \int_{0}^{\pi }{\frac{x(2sinxcosx)}{1- sin^{2}xcos^{2}x}}dx

I_{1}=\int_{0}^{\pi }{\frac{\pi(2sinxcosx)}{1- sin^{2}xcos^{2}x}}dx + I_{1}

Yahin pe mein atak gaya hoon. Please koi meri galti point out kar do.

@Che : Ya I got the same ans

But plzz post ure method also

I want to learn some easier of solving this one ??

i actually got teh soln...not actually mine's but someone else's

ans is \frac{7\pi^{2}}{6\sqrt{3}} $

uttara check if u r getting teh same ans with ur method....i din check it

2I = \int_{0}^{\Pi }{x\left[2/(1-sin^{2}xcos^{2}x \right]} dx + \Pi \int_{0}^{\Pi }{1/\left(1-sinxcosx \right)}dx

I_{1} = \int_{0}^{\Pi }{x\left[(-2sinxcosx)/(1-sin^2xcos^2x) \right]} dx

3I_{1} = \Pi \int_{0}^{\Pi }{(2sinxcosx)/(1-sin^2xcos^2x)} dx

I_{2}=\Pi \int_{0}^{\Pi }{1/\left(1-sinxcosx \right)}dx = % I_{2} = \int_{0}^{\Pi }{(sec^2x)/(sec^2x - tanx)} dx
I_{2} = \int_{0}^{\Pi }{(sec^2x)/(sec^2x - tanx)} dx

Put tanx = t
sec^2x dx = dt

Now solve I 2

∫1/[(t-1/2)² +(√3/2)²

Now replace I1 & I 2 in the I eqn and solve for sec^4 x using the formulae

To get the first step i replaced x by pi-x & added both

Repeated the same method for I 1 also

ya....I know tht the ans is wrong.
......som1 pls point out the mistake..........

Another try...................
I=2x/(1+(sinx+cosx)²)
I=2x/(1+(cos^{2}x/2-sin^{2}x/2+2sinx/2.cosx/2)^{2}

2I=2\int_{0}^{\pi }{}
\pi /(1+(cos^{2}x/2-sin^{2}x/2+2sinx/2.cosx/2)^{2}

I=\int_{0}^{\pi }{\pi /(1+(sinx+cosx)^{2})}=\int_{0}^{\pi }{\pi /(2+sin2x)}

hmm....thts rong.........

@ Rahul : How u got that 1st step ??