if you want it telescopic...

Calculus Level 5

Given: $\displaystyle \int_{0}^1 \tan^{-1} (1-x+x^2) \ dx =a\pi+\log_e(\lambda)$

Find the value of $a-\lambda$ .

Try more Integral problems here .

The answer is -2.

1 solution

Abhishek Maji
Nov 24, 2014

$\displaystyle \int_{0}^1 tan^{-1}(1-x+x^2).dx=\int_{0}^1 \frac{\pi}{2}-cot^{-1}(1-x+x^2).dx$

And $\int_{0}^1 cot^{-1}(1-x+x^2).dx$ is solved in the previous problem.. You can check it here

So $\int_{0}^1 tan^{-1}(1-x+x^2).dx=log_e2$

From here, we get $a=0, \lambda=2$

$\implies a-\lambda=\boxed{-2}$ .

see this , here the answer comes as $\frac{\pi}{2} - ln2$

so according to your answer its integral should be $\frac{\pi}{2} - (\frac{\pi}{2} - ln2) = ln2$

thus comparing we get a=0 and $\lambda = \frac{1}{2}$

so the answer should be -0.5 @Abhishek Maji

sandeep Rathod - 6 years, 6 months ago

I think you are correct @sandeep Rathod

Abhishek Maji - 6 years, 6 months ago

Please post your comment on this @Sandeep Bhardwaj

Abhishek Maji - 6 years, 6 months ago

@Abhishek Maji – You are right guys. I am extremely sorry for this silly mistake. I am editing the problem according to the answer. I make you sure it not to happen in the future. I will be very careful. Thanks for observing it. @sandeep Rathod @Abhishek Maji

Sandeep Bhardwaj - 6 years, 6 months ago

@Abhishek Maji – I am editing your solution. Please post your solutions with your good latex. Its my humble request to you. @Abhishek Maji

Sandeep Bhardwaj - 6 years, 6 months ago

No . lambda is 2.also! this is a 1998 jee problem!!!!!

Rushikesh Joshi - 6 years, 6 months ago

if you want it telescopic...

The answer is -2.

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