integration

integrate (tan^-1)^2

Note by Sriram Raghavan
7 years, 9 months ago

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Comments

Is this (arctan⁡(x))2(\arctan(x))^2 or (cot⁡(x))2(\cot(x))^2

Edward Jiang - 7 years, 9 months ago

You mean (arctanx)2(arctan x)^2, right?

Shourya Pandey - 7 years, 9 months ago

its tan inverse x whole square

Sriram Raghavan - 7 years, 9 months ago

use of integration in daily life

meerab jutt - 7 years, 8 months ago

Given ∫(tan⁡−1(x))2dx\displaystyle \int \left(\tan^{-1}(x)\right)^2dx

Now Let tan⁡−1(x)=t⇔x=tan⁡(t)⇔dx=sec⁡2(t)dt\tan^{-1}(x) = t\Leftrightarrow x = \tan (t)\Leftrightarrow dx = \sec^2 (t) dt

So ∫t2⋅sec⁡2(t)dt\displaystyle \int t^2 \cdot \sec^2 (t) dt

Now Using Integration by parts , we get

=t2tan⁡(t)−2∫t⋅tan⁡(t)dt\displaystyle =t^2 \tan (t)-2 \int t\cdot \tan (t)dt

Again using I.B.P, we get

=t2⋅tan⁡(t)−2t⋅ln⁡∣sec⁡(t)∣+2∫ln⁡∣sec⁡(t)∣dt\displaystyle = t^2 \cdot \tan (t)-2t\cdot \ln \left|\sec (t)\right|+2\int \ln \left|\sec (t)\right| dt

Now I did not understand How can i solve ∫ln⁡∣sec⁡(t)∣dt\displaystyle \int \ln \left|\sec (t)\right|dt

OR may be yours question is like ∫0π4(tan⁡−1(x))2dx\displaystyle \int_{0}^{\frac{\pi}{4}}\left(\tan^{-1}(x)\right)^2dx

OR ∫0π2(tan⁡−1(x))2dx\displaystyle \int_{0}^{\frac{\pi}{2}}\left(\tan^{-1}(x)\right)^2dx

jagdish singh - 7 years, 8 months ago
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