My second integral problem

Calculus Level 3

Evaluate $\displaystyle \int_{0}^{\frac{3}{5}} \frac{\text{d} x}{\sqrt{\left(1-x^{2}\right)^{3}}}$ . If the answer can be put in the form $\dfrac{a}{b}$ , where $a$ and $b$ are coprime, find $a+b$ ..

The answer is 7.

1 solution

Brian Charlesworth
Dec 13, 2015

Let $x = \sin(\theta).$ Then $dx = \cos(\theta) d\theta$ and $1 - x^{2} = \cos^{2}(\theta).$

Also, as $x$ goes from $0$ to $\frac{3}{5}$ we have $\theta$ going from $0$ to $\sin^{-1}(\frac{3}{5}).$ On this interval $\cos(\theta)$ is positive, so we have that $\sqrt{(1 - x^{2})^{3}} = \sqrt{\cos^{6}(\theta)} = \cos^{3}(\theta).$ The integral thus becomes

$\displaystyle\int_{0}^{\sin^{-1}(\frac{3}{5})} \dfrac{\cos(\theta)}{\cos^{3}(\theta)} d\theta = \int_{0}^{\sin^{-1}(\frac{3}{5})} \sec^{2}(\theta) d\theta = \tan(\sin^{-1}(\frac{3}{5})) - \tan(0) = \frac{3}{4}.$

The desired solution is then $a + b = 3 + 4 = \boxed{7}.$

how did you calculate tan(sen^-1(3/5))? please

guido barta - 5 years, 5 months ago

Form a right triangle $\Delta ABC$ with $\angle ABC = 90^{\circ}, |BC| = 3$ and $|AC| = 5.$

Then $\sin(\angle CAB) = \dfrac{3}{5} \Longrightarrow \angle CAB = \sin^{-1}(\frac{3}{5})$ . Also, $|AB| = \sqrt{|AC|^{2} - |BC|^{2}} = 4.$

So $\tan(\sin^{-1}(\frac{3}{5})) = \tan(\angle CAB) = \dfrac{|BC|}{|AB|} = \dfrac{3}{4}.$

Brian Charlesworth - 5 years, 5 months ago

thanks brian see u soon on brilliant

guido barta - 5 years, 5 months ago

My second integral problem

The answer is 7.

1 solution

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