Question

Solve the integral of √(49-x²)÷x from ln1 to ln2

Answer 1

$\int_{ln1}^{ln2} {\frac{\sqrt{49-x²}}{x}}dx$

Let x = 7sin$\theta$

dx = 7cos$\theta$ d$\theta$

So $\int {\frac{\sqrt{49-x²}}{x}}dx$ =

$\int {\frac{\sqrt{49-49sin²\theta}}{7sin\theta}}7cos{\theta}d\theta$

= $\int {\frac{7cos²{\theta}}{sin\theta}}d\theta$

= $7\int {\frac{1-sin²{\theta}}{sin\theta}}d\theta$

= $7\int [{cosec\theta+ sin\theta}]d\theta$

Let ln(cosec$\theta$ + cot$\theta$) = t

So $\frac{-cosec\theta cot\theta-cosec²\theta)}{cosec\thetaθ + cot\thetaθ)}d\theta$= dt

=> cosec$\theta$ d$\theta = -dt$

= $-7\int dt -7\int sin\theta d\theta$

= -7t + 7cos$\theta$ + C

= -7 ln|cosec$\theta$ + cot$\theta$ | + 7cos$\theta$

= 7 ln| $\frac{sin\theta}{1+cos\theta}$ | + 7cos$\theta$ + C

= $ln[\frac{2sin\theta/2cos\theta/2}{2cos²\theta/2}]+ 7cos\theta + C$

= ln[tan $\theta/2]+ 7cos\theta+C$

Now as x--> ln1 , $\theta$ --> 0 and as x--> ln2 , $\theta$ --> sin ^-1(ln2/7)

So

$\int_{ln1}^{ln2} {\frac{\sqrt{49-x²}}{x}}dx$ = $[ln(tan\theta/2)+7cos\theta]]_{0}^{sin^{-1}(ln2/7)}$

= $[ln(tan(\frac{1}{2}sin^{-1}(ln(2)/7)+$

$7cos((sin^{-1}(ln(2/7)]-$

$[ln(tan(0)+$ $7cos(0)]$

= $[ln(tan(\frac{1}{2}sin^{-1}(ln(2)/7)+$

$7cos((sin^{-1}(ln(2/7)] -7 - ln(0)$

= A finite quantity + ∞ as ln(0) -->-∞

= ∞