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Grade 12Integral Calculus

integrate the following : sin-1 x/a+x the whole raised to 1/2.

Profile image of sahuh
8 Years agoGrade 12
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2 Answers

Profile image of Arun
8 Years ago
 

Dear student,

 

Make the substitution, x = a*tan2θ

So dx = 2a*tanθ*sec2θ dθ

and x/(a+x) = sin2θ

 

∫θ*2a*tanθ*sec2θ dθ = 2a∫θ secθ*(secθtanθ) dθ.

Integrate this by parts

= a*[θsec2θ - ∫sec2θdθ] = a[θsec2θ - tanθ] + constant.

where sec2θ = 1 + (x/a), and tanθ = √(x/a), θ = tan-1√(x/a)

 

Hope that helps.

 

 

Regards

Arun (askIITians forum expert)

Profile image of Arun
8 Years ago
Dear student,
 
Also you can solve it by another method which is as follows-
 
 
 
I = \int sin^{-1}\sqrt{\frac{x}{x+a}}dx
Integration by Parts
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\int \frac{1}{2}\sqrt{\frac{ax}{(x+a)^{2}}}dx
t = \sqrt{x}
dt = \frac{1}{2\sqrt{x}}dx
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}\int\frac{t^{2}}{a+t^{2}}dt
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}t +atan^{-1}(\frac{t}{\sqrt{a}}) +constant
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}\sqrt{x}+atan^{-1}(\frac{\sqrt{x}}{\sqrt{a}}) +constant
 
 
Hope it helps.
 
 
 
Regards
Arun

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