Question

Evaluate the following definite integrals by finding antiderivatives : (i) \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}}\) (ii) \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}(b>a)\) (iii) \(\int_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}\) (iv) \(\int_{2}^{3} \frac{\sqrt{(x-2)^{2}}}{1+\sqrt{(x-2)^{3}}} d x\)

Short answer

Question: Evaluate the following integrals: (i) \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}}\) (ii) \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}\) (iii) \(\int_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}\) (iv) \(\int_{2}^{3} \frac{\sqrt{(x-2)^{2}}}{1+\sqrt{(x-2)^{3}}} dx\) Answer: (i) \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}} = 1\) (ii) \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}= 2\left(\frac{\pi}{2}-\sin^{-1}{\sqrt\frac{a-\alpha}{\beta-\alpha}}\right)\) (iii) \(\int_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}= \frac{\pi}{10} - \frac{\ln 4}{5}\) (iv) \(\int_{2}^{3} \frac{\sqrt{(x-2)^{2}}}{1+\sqrt{(x-2)^{3}}} dx = \frac{3}{2}\)

Step-by-step solution

(i) Integration of the First Function

The first integral is \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}}\). To solve this integral, we first notice that we can perform a substitution. Let \(x = \frac{1}{\sqrt{2}}\sin{u}\). Then, \(dx = \frac{1}{\sqrt{2}}\cos{u} du\). Now, we can substitute this into our integral: \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}} = \int_{0}^{\pi/4} \frac{\frac{1}{\sqrt{2}}\cos{u} du}{\left(1 - 2\left(\frac{1}{\sqrt{2}}\sin{u}\right)^{2}\right) \sqrt{1-\left(\frac{1}{\sqrt{2}}\sin{u}\right)^{2}}}\). Solving this integral, we get: \(\int_{0}^{\pi/4} \frac{\cos{u} du}{\cos^{2}{u}} = \int_{0}^{\pi/4} \sec^{2}{u} du\). Now, we know that the antiderivative of \(\sec^{2}{u}\) is \(\tan{u}\). Therefore, we can now evaluate the integral: \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}} = \left[\tan{u}\right]_{0}^{\pi/4} = \left[\tan(\frac{\pi}{4}) - \tan(0)\right]\). Finally, we find the answer to be: (i) \(\int_{0}^{1 / 2} \frac{d x}{\left(1-2 x^{2}\right) \sqrt{1-x^{2}}} = 1 - 0 = 1\).

(ii) Integration of the Second Function

The second integral is \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}\). To solve this integral, we perform a substitution. Let's use the following substitution: \(x = \alpha + (\beta - \alpha) \sin^2{u},\) and \(dx = 2(\beta - \alpha) \sin{u}\cos{u} du\). Now, we can substitute this into our integral: \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}} = \int_{\sin^{-1}{\sqrt\frac{a-\alpha}{\beta-\alpha}}}^{\pi/2} \frac{2(\beta - \alpha) \sin{u}\cos{u} du}{\sqrt{(\beta - \alpha) \sin^{2}{u}(\beta - \alpha)(1 - \sin^{2}{u})}}\). Solving this integral, we get: \(\int_{\sin^{-1}{\sqrt\frac{a-\alpha}{\beta-\alpha}}}^{\pi/2} 2 du\). Now we evaluate the integral and get the answer: (ii) \(\int_{a}^{\beta} \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}= 2\left[\cancel{u}\right]^{\pi/2}_{\sin^{-1}{\sqrt\frac{a-\alpha}{\beta-\alpha}}}=2\left(\frac{\pi}{2}-\sin^{-1}{\sqrt\frac{a-\alpha}{\beta-\alpha}}\right)\).

(iii) Integration of the Third Function

The third integral is \(\int_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}\). To solve this integral, we use the substitution: \(u = \cot{x} \Rightarrow du = -\csc^2{x} dx\) Now, we can substitute this into our integral: \(\int_{0}^{\pi / 2} \frac{dx}{4+5 \sin x} = \int_{\infty}^{0} \frac{-du}{4+5 (1 - \csc^2{u})}\) Solving this integral, we get: \(\int_{\infty}^{0} -\frac{du}{4+5 - 5\csc^2{u}} = \int_{\infty}^{0} \frac{du}{(u^2+1)+4u}\) Now, we can use partial fractions to simplify the integral: \(\frac{5}{(u^2+1)(u+4)} = \frac{A}{u^2+1} + \frac{B}{u+4}\) Solving for A and B, we get \(A=\frac{1}{5}\) and \(B=-\frac{1}{5}\). So the integral becomes: \(\int_{\infty}^{0} \frac{1}{5}\frac{du}{u^2+1} - \frac{1}{5}\frac{du}{u+4}\) Now, we integrate and get the answer: (iii) \(\int_{0}^{\pi / 2} \frac{d x}{4+5 \sin x}= \left[\frac{1}{5}\tan^{-1}{u}\right]_{\infty}^{0}-\left[\frac{1}{5}\ln|u+4|\right]_{\infty}^{0}=\frac{\pi}{10} - \frac{\ln 4}{5}\).

(iv) Integration of the Fourth Function

The fourth integral is \(\int_{2}^{3} \frac{\sqrt{(x-2)^{2}}}{1+\sqrt{(x-2)^{3}}} dx\). Let's simplify the expression first: \(\int_{2}^{3} \frac{\sqrt{(x-2)^2}}{1+\sqrt{(x-2)^3}}dx = \int_{2}^{3} \frac{x-2}{1+\sqrt{(x-2)^3}}dx\) Now, we can use the substitution: \(u = (x-2)^3 \Rightarrow u^{1/3} = (x-2) \Rightarrow x = u^{1/3}+2\) and \(dx = \frac{1}{3}u^{-2/3} du\) Now, we substitute this into our integral: \(\int_{2}^{3} \frac{x-2}{1+\sqrt{(x-2)^3}}dx = \int_{0}^{1} \frac{1}{1+u^{1/3}}\frac{1}{3}u^{-2/3} du\) Simplifying the integral, we get: \(\int_{0}^{1} \frac{1}{3(1+u^{1/3})}\frac{1}{u^{2/3}} du = \frac{1}{3}\int_{0}^{1} \frac{du}{(1+u^{1/3})u^{2/3}}\) Now, we integrate and get the answer: (iv) \(\int_{2}^{3} \frac{\sqrt{(x-2)^{2}}}{1+\sqrt{(x-2)^{3}}} dx = \left[\frac{3}{2}u^{1/3}\right]_{0}^{1}=\frac{3}{2}(1^{1/3}-0) = \frac{3}{2}\).