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# integration

integrate (tan^-1)^2

Note by Sriram Raghavan
3 years, 6 months ago

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

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

So $$\displaystyle \int t^2 \cdot \sec^2 (t) dt$$

Now Using Integration by parts , we get

$$\displaystyle =t^2 \tan (t)-2 \int t\cdot \tan (t)dt$$

Again using I.B.P, we get

$$\displaystyle = t^2 \cdot \tan (t)-2t\cdot \ln \left|\sec (t)\right|+2\int \ln \left|\sec (t)\right| dt$$

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

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

OR $$\displaystyle \int_{0}^{\frac{\pi}{2}}\left(\tan^{-1}(x)\right)^2dx$$ · 3 years, 6 months ago

use of integration in daily life · 3 years, 6 months ago

its tan inverse x whole square · 3 years, 6 months ago

You mean $$(arctan x)^2$$, right? · 3 years, 6 months ago

Is this $(\arctan(x))^2$ or $(\cot(x))^2$ · 3 years, 6 months ago