int_0^1 {{tan ^{ - 1}}x,dx = } | vedclass

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(A) Let $I = \int_0^1 {{	an ^{ - 1}}x\,dx}$.Using integration by parts,$\int {u,dv = uv - \int {v,du} } $.Let $u = {	an ^{ - 1}}x$ and $dv = dx$.The

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$\frac{\pi }{4} - \frac{1}{2}\log 2$

Vedclass · Answer

(A) Let $I = \int_0^1 {{	an ^{ - 1}}x\,dx}$.Using integration by parts,$\int {u,dv = uv - \int {v,du} } $.Let $u = {	an ^{ - 1}}x$ and $dv = dx$.Then $du = \frac{1}{{1 + {x^2}}}dx$ and $v = x$.$I = [x{	an ^{ - 1}}x]_0^1 - \int_0^1 {\frac{x}{{1 + {x^2}}}dx} $.$I = (1 \cdot {	an ^{ - 1}}(1) - 0 \cdot {	an ^{ - 1}}(0)) - \frac{1}{2}\int_0^1 {\frac{{2x}}{{1 + {x^2}}}dx} $.$I = \frac{\pi }{4} - \frac{1}{2}[\log (1 + {x^2})]_0^1$.$I = \frac{\pi }{4} - \frac{1}{2}(\log 2 - \log 1)$.Since $\log 1 = 0$,we get $I = \frac{\pi }{4} - \frac{1}{2}\log 2$.

$\int_0^1 {{\tan ^{ - 1}}x\,dx = } $

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