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.