How can you proof int dx/(a^2-x^2)=1/(2a) log |(a+x)/(a-x)| + cdxa2x2=12aloga+xax+c using x = a sinthetax=asinθ ?

int dx/(a^2-x^2)=1/(2a) log |(a+x)/(a-x)| + cdxa2x2=12aloga+xax+c

using x = a sinthetax=asinθ

3 Answers
Jul 19, 2018

Kindly refer to Explanation.

Explanation:

Suppose that, I=int1/(a^2-x^2)dxI=1a2x2dx.

Using the substn., x=asintheta, so, dx=acosthetad thetax=asinθ,so,dx=acosθdθ,

I=int(acostheta)/(a^2-a^2sin^2theta)d thetaI=acosθa2a2sin2θdθ,

=int(acostheta)/(a^2cos^2theta)d theta=acosθa2cos2θdθ,

=1/aint1/costhetad theta=1a1cosθdθ,

=1/aintsecthetad theta=1asecθdθ,

=1/aln|(sectheta+tantheta)|=1aln|(secθ+tanθ)|,

=1/aln|(1/costheta+sintheta/costheta)|=1aln(1cosθ+sinθcosθ),

=1/aln|(1+sintheta)/costheta|=1aln1+sinθcosθ,

=1/(2a){2ln|(1+sintheta)/costheta|}=12a{2ln1+sinθcosθ},

=1/(2a){ln|(1+sintheta)/costheta|^2}=12a{ln1+sinθcosθ2},

=1/(2a)ln|(1+sintheta)^2/cos^2theta|=12aln∣ ∣(1+sinθ)2cos2θ∣ ∣,

=1/(2a)ln|(1+sintheta)^2/(1-sin^2theta)|=12aln∣ ∣(1+sinθ)21sin2θ∣ ∣,

=1/(2a)ln|(1+sintheta)^2/((1+sintheta)(1-sintheta))|=12aln∣ ∣(1+sinθ)2(1+sinθ)(1sinθ)∣ ∣,

=1/(2a)ln|(1+sintheta)/(1-sintheta)|=12aln1+sinθ1sinθ,

=1/(2a)ln|(a+asintheta)/(a-asintheta)|=12alna+asinθaasinθ,

=1/(2a)ln|(a+x)/(a-x)|+c=12alna+xax+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=asinthetax=asinθ, =>, dx=a costhetad thetadx=acosθdθ

a^2-x^2=a^2-a^2sin^2theta=a^2cos^2thetaa2x2=a2a2sin2θ=a2cos2θ

Therefore,

The integral is

I=int(dx)/(a^2-x^2)=int(acosthetad theta)/(a^2cos^2theta)=1/aintsecthetad thetaI=dxa2x2=acosθdθa2cos2θ=1asecθdθ

=1/aint(sectheta(tantheta+sectheta)d theta)/(tantheta+sectheta)=1asecθ(tanθ+secθ)dθtanθ+secθ

=1/aint((secthetatantheta+sec^2theta)d theta)/(tantheta+sectheta)=1a(secθtanθ+sec2θ)dθtanθ+secθ

Let u=tantheta+secthetau=tanθ+secθ

=>, du=(secthetatantheta+sec^2theta)d thetadu=(secθtanθ+sec2θ)dθ

Therefore,

I=1/aint(du)/uI=1aduu

=1/aln(u)=1aln(u)

=1/aln(tantheta+sectheta)=1aln(tanθ+secθ)

=1/aln(a/(sqrt(a^2-x^2))+x/(sqrt(a^2-x^2)))=1aln(aa2x2+xa2x2)

=1/aln(a+x)-1/aln(sqrt(a^2-x^2))=1aln(a+x)1aln(a2x2)

=1/aln(a+x)-1/(2a)ln(a^2-x^2)=1aln(a+x)12aln(a2x2)

=2/(2a)ln(a+x)-1/(2a)ln(a^2-x^2)=22aln(a+x)12aln(a2x2)

=1/2a(ln((a+x)^2/(a^2-x^2)))=12a(ln((a+x)2a2x2))

=1/2aln(((a+x)(a+x))/((a+x)(a-x)))=12aln((a+x)(a+x)(a+x)(ax))

=1/(2a)ln|(a+x)/(a-x)| + C=12alna+xax+C

QEDQED

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