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integrate (tan^-1)^2

Note by Sriram Raghavan 4 years, 8 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 \(\displaystyle \int_{0}^{\frac{\pi}{2}}\left(\tan^{-1}(x)\right)^2dx\)

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use of integration in daily life

its tan inverse x whole square

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

Is this \[(\arctan(x))^2\] or \[(\cot(x))^2\]

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## Comments

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

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 \(\displaystyle \int_{0}^{\frac{\pi}{2}}\left(\tan^{-1}(x)\right)^2dx\)

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use of integration in daily life

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its tan inverse x whole square

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

Log in to reply

Is this \[(\arctan(x))^2\] or \[(\cot(x))^2\]

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