InterviewSolution
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InterviewSolution
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`int_(-a)^asqrt((a-x)/(a+x))dx` |
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Answer» Put `x=acostheta ` so that `dx=-asin theta d theta`. Also, `(x=-aimpliestheta=pi)` and `(x=aimpliestheta=0)`. `:.int_(-a)^(a)sqrt((a-x)/(a+x))dx=int_(pi)^(0)sqrt((1-costheta)/(1+costheta))*(-asintheta)d theta` `=aint_(0)^(pi)sqrt((2sin^(2)(theta//2))/(2cos^(2)(theta//2)))*2sin(theta//2)cos(theta//2)d theta` `=a int_(0)^(pi)2sin^(2)(theta//2)d theta=a int_(0)^(pi)(1-costheta)d theta` `=a int_(0)^(pi)d theta-a int_(0)^(pi)cos theta d theta` `=a*[theta]_(0)^(pi)-a[sintheta]_(0)^(pi)=api`. |
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