INTEGRAL

integration by parts will be still messy qwerty can u integrate this
∫tan^-1(2tan+1/root3)?????

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}}}

No one's able to get the ans. !!

@Che : Ya I got the same ans

But plzz post ure method also

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

A request to all users:- please do not delete your posts if you have wrongly attempted them. You may use these post for your future reference to identify where you went wrong in your first attempt. This will prove you to be really helpful in identifying your mistakes. Plus, this discussion helps other users to understand what mistakes should not be done.

I meant no offence. Just wanted to help all my fellow tiitians.

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

Again I made a mistake but here is the correction

I1 ≠0

3 I1 = pi ∫[(2sc)/(1 - s^2 c^2 ) dx

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

Repeated the same method for I 1 also

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

I1 = \int_{0}^{\Pi }{x[(-2sinxcosx)/(1-sin^2xcos^2x)]} = \int_{0}^{\Pi }{x[(2sinxcosx)/(1-sin^2xcos^2x)]}

2 I_{1} = 0

I_{2}=\Pi \int_{0}^{\Pi }{1/\left(1-sinxcosx \right)}dx = \Pi \int_{0}^{\Pi }{\left[1/(1-sin^{2}xcos^{2}x)] \right]} dx = {} ∫ [0 t0 pi ] sec^4x

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

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

sorry sorry ... realised it when i tried it myself .. lolz [5]

u hav x in the numerator !

ans is 6 .299

=(\pi/√3)\tan^{-1} (2\pi +1/\sqrt{3})-\pi /6\sqrt{3}

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.........

can u jus explain how u got ur I in post 9 i mean the denominater part

thats rong

how does that help ? btw that does not equal to the denominater

ya how u got ur 1st steprahul ?