How do you integrate $int x /sqrt(1 + x^2) dx$ using trigonometric substitution?

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Answer 1 · 2016-06-04T03:23:07

$int x/sqrt(1+x^2) dx=sqrt(1+x^2)+C$

Integrate $int x/sqrt(1+x^2) dx$ by Trigonometric substitution

Solution:

Let $x=tan theta$
Let $x^2=tan^2 theta$
Let $dx=sec^2 theta* d theta$

$int x/sqrt(1+x^2) dx=int (tan theta*sec^2 theta* d theta)/sqrt(1+tan^2 theta)=int (tan theta*sec^2 theta* d theta)/sqrt(sec^2 theta)$

$int x/sqrt(1+x^2) dx=int (tan theta*sec^2 theta* d theta)/sec theta$

$int x/sqrt(1+x^2) dx=int tan theta*sec theta* d theta$

$int x/sqrt(1+x^2) dx=int (sin theta/cos theta*1/cos theta)* d theta$

$int x/sqrt(1+x^2) dx=int sin theta/cos^2 theta d theta$

$int x/sqrt(1+x^2) dx=int cos^(-2) theta*sin theta* d theta$

$int x/sqrt(1+x^2) dx=-int cos^(-2) theta*(-sin theta)* d theta$

$int x/sqrt(1+x^2) dx=- (cos theta)^(-2+1)/(-2+1)$

$int x/sqrt(1+x^2) dx=(cos theta)^(-1)=sec theta$

Our Right Triangle with acute angle $theta$

$tan theta=x/1$

$sec theta=sqrt(1+x^2)/1$

$int x/sqrt(1+x^2) dx=(cos theta)^(-1)=sec theta=sqrt(1+x^2)/1$

Therefore

$int x/sqrt(1+x^2) dx=sqrt(1+x^2)+C$

God bless....I hope the explanation is useful.

How do you integrate int x /sqrt(1 + x^2) dxint x /sqrt(1 + x^2) dx using trigonometric substitution?