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
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!

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

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