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Evaluate – > ∫ tan(ln( x + (1+x 2 ) 1/2 )) / ( 1+ x 2 ) 1/2 dx

Evaluate – > ∫  tan(ln( x + (1+x2)1/2)) / ( 1+ x2)1/2 dx

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
I = \int \frac{ tan(ln(x+\sqrt{1+x^{2}}))}{\sqrt{1+x^{2}}}dx
x = tan\theta
dx = sec^{2}\theta.d\theta
I = \int \frac{ tan(ln(tan\theta +\sqrt{1+tan^{2}\theta }))}{\sqrt{1+tan^{2}\theta }}.sec^{2}\theta .d\theta
I = \int { tan(ln(tan\theta + sec\theta )).sec\theta.d\theta
(ln(tan\theta + sec\theta )) = t
\frac{sec^{\2 }\theta +sec\theta .tan\theta }{tan\theta +sec\theta }.d\theta = dt
sec\theta .d\theta =dt
I = \int tant.dt
I = ln|sect| + c
I = ln|sec(ln(tan\theta +sec\theta ))| + c
I = ln|sec(ln(x +\sqrt{1+x^{2}} ))| + c
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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