Question #03e6a

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Question #03e6a

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Answer 1 · 2017-04-12T23:59:25

$= \frac{1}{2} ({tan}^{- 1} x + \frac{x}{1 + x^{2}})$ $=1/2 (tan^(-1) x+x/(1+x^2))$

Explanation:

$\int_{0}^{{tan}^{- 1} x} \frac{1}{1 + {tan}^{2} t} d t$ $int_0^(tan^(-1) x)1/(1+tan^2 t) dt$

By Trigonometric Identities,

$= \int_{0}^{{tan}^{- 1} x} \frac{1}{{sec}^{2} t} d t = \int_{0}^{{tan}^{- 1} x} {cos}^{2} t d t = \int_{0}^{{tan}^{- 1} x} \frac{1}{2} (1 + cos 2 t) d t$ $=int_0^(tan^(-1) x)1/sec^2 t dt =int_0^(tan^(-1) x)cos^2 t dt =int_0^(tan^(-1)x)1/2(1+cos 2t)dt$

By integrating and $sin 2 t = 2 sin t cos t$ $sin 2t=2sin t cos t$

$= \frac{1}{2} {[1 + \frac{sin 2 t}{2}]}_{0}^{{tan}^{- 1} x} = \frac{1}{2} {[t + sin t cos t]}_{0}^{{tan}^{- 1} x}$ $=1/2 [1+(sin 2t)/2]_0^(tan^(-1) x) =1/2[t+sin t cos t]_0^(tan^(-1) x)$

$= \frac{1}{2} ({tan}^{- 1} x + sin ({tan}^{- 1} x) \cdot cos ({tan}^{- 1} x))$ $=1/2(tan^(-1) x +sin(tan^(-1) x) cdot cos(tan^(-1) x))$

$= \frac{1}{2} ({tan}^{- 1} x + \frac{x}{\sqrt{1 + x^{2}}} \cdot \frac{1}{\sqrt{1 + x^{2}}})$ $=1/2(tan^(-1) x+ x/sqrt(1+x^2) cdot 1/sqrt(1+x^2))$

$= \frac{1}{2} ({tan}^{- 1} x + \frac{x}{1 + x^{2}})$ $=1/2(tan^(-1) x+x/(1+x^2))$

I hope that this was clear.

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Question #03e6a

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