How can you proof #int dx/(a^2-x^2)=1/(2a) log |(a+x)/(a-x)| + c# using #x = a sintheta# ?

#int dx/(a^2-x^2)=1/(2a) log |(a+x)/(a-x)| + c#

using #x = a sintheta#

3 Answers
Ratnaker Mehta
Jul 19, 2018

Kindly refer to Explanation.

Explanation:

Suppose that, #I=int1/(a^2-x^2)dx#.

Using the substn., #x=asintheta, so, dx=acosthetad theta#,

#I=int(acostheta)/(a^2-a^2sin^2theta)d theta#,

#=int(acostheta)/(a^2cos^2theta)d theta#,

#=1/aint1/costhetad theta#,

#=1/aintsecthetad theta#,

#=1/aln|(sectheta+tantheta)|#,

#=1/aln|(1/costheta+sintheta/costheta)|#,

#=1/aln|(1+sintheta)/costheta|#,

#=1/(2a){2ln|(1+sintheta)/costheta|}#,

#=1/(2a){ln|(1+sintheta)/costheta|^2}#,

#=1/(2a)ln|(1+sintheta)^2/cos^2theta|#,

#=1/(2a)ln|(1+sintheta)^2/(1-sin^2theta)|#,

#=1/(2a)ln|(1+sintheta)^2/((1+sintheta)(1-sintheta))|#,

#=1/(2a)ln|(1+sintheta)/(1-sintheta)|#,

#=1/(2a)ln|(a+asintheta)/(a-asintheta)|#,

#=1/(2a)ln|(a+x)/(a-x)|+c#, as desired!

Narad T.
Jul 19, 2018

Please see the proof below.

Explanation:

Normally, you perform this integral by the composition into partial fractions.

But you need.

#x=asintheta#, #=>#, #dx=a costhetad theta#

#a^2-x^2=a^2-a^2sin^2theta=a^2cos^2theta#

Therefore,

The integral is

#I=int(dx)/(a^2-x^2)=int(acosthetad theta)/(a^2cos^2theta)=1/aintsecthetad theta#

#=1/aint(sectheta(tantheta+sectheta)d theta)/(tantheta+sectheta)#

#=1/aint((secthetatantheta+sec^2theta)d theta)/(tantheta+sectheta)#

Let #u=tantheta+sectheta#

#=>#, #du=(secthetatantheta+sec^2theta)d theta#

Therefore,

#I=1/aint(du)/u#

#=1/aln(u)#

#=1/aln(tantheta+sectheta)#

#=1/aln(a/(sqrt(a^2-x^2))+x/(sqrt(a^2-x^2)))#

#=1/aln(a+x)-1/aln(sqrt(a^2-x^2))#

#=1/aln(a+x)-1/(2a)ln(a^2-x^2)#

#=2/(2a)ln(a+x)-1/(2a)ln(a^2-x^2)#

#=1/2a(ln((a+x)^2/(a^2-x^2)))#

#=1/2aln(((a+x)(a+x))/((a+x)(a-x)))#

#=1/(2a)ln|(a+x)/(a-x)| + C#

#QED#

Harish Chandra Rajpoot
Jul 19, 2018

Proof is below

Explanation:

Let #x=a\sin\theta\implies dx=a\cos\theta\ d\theta#

#\therefore \int \frac{dx}{a^2-x^2}#

#=\int \frac{a\cos\theta\ d\theta}{a^2-a^2\sin^2\theta}#

#=\int \frac{a\cos\theta\ d\theta}{a^2\cos^2\theta}#

#=1/a\int \frac{\ d\theta}{\cos \theta}#

#=1/a\int \sec \theta \ d\theta#

#=1/a\int {\sec \theta(\sec \theta+\tan\theta)}/{\sec\theta+\tan\theta} \ d\theta#

#=1/a\int {(\sec \theta\tan\theta+\sec^2\theta)\ d\theta}/{\sec\theta+\tan\theta}#

#=1/a\int {d(\sec \theta+tan\theta)}/{\sec\theta+\tan\theta}#

#=1/a\ln|\sec\theta+\tan\theta|+c#

#=1/a\ln|1/{\cos\theta}+{\sin\theta}/{\cos\theta}|+c#

#=1/a\ln|{1+\sin\theta}/{\cos\theta}|+c#

#=1/a\ln|{1+x/a}/{\sqrt{1-\sin^2\theta}}|+c#

#=1/a\ln|{{a+x}/a}/{\sqrt{1-x^2/a^2}}|+c#

#=1/a\ln|{a+x}/(a^2-x^2)^{1/2}|+c#

#=1/a\cdot 1/2 \ln|{(a+x)^2}/{a^2-x^2}|+c#

#=1/{2a} \ln|{(a+x)^2}/{(a+x)(a-x)}|+c#

#=1/{2a}\ln|{a+x}/{a-x}|+c#

Related questions
Impact of this question
14070 views around the world
You can reuse this answer
Creative Commons License