Question

Evaluate the following definite integrals by finding antiderivatives: (i) \(\int_{3}^{8} \frac{\sin \sqrt{x+1}}{\sqrt{x+1}} d x\) (ii) \(\int_{0}^{\pi / 4} \cos 2 x \sqrt{4-\sin 2 x} d x\) (iii) \(\int_{0}^{1-e^{-2}} \frac{\ln (1-t)}{1-t} \mathrm{dt}\) (iv) \(\int_{0}^{1} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)

Short answer

Short answer: (i) \(-\cos(3)+\cos(2)\) (ii) \(\frac{2}{3}(3^{3/2}-4^{3/2})\) (iii) Since it is difficult to find an elementary antiderivative for the remaining integral, we leave the answer as: \(\int_{0}^{1-e^{-2}}(1-t)\ln(1-t)\, dt\) (iv) \(\frac{\pi}{2}\)

Step-by-step solution

Substitution

Let \(u = \sqrt{x+1}\). Then, we have \(u^2 = x+1\) and \(2u\,du = dx\). The integral becomes $$\int_{u(3)}^{u(8)} \frac{\sin(u)}{u}(2u\, du).$$

Calculate the new limits

When \(x = 3\), \(u = \sqrt{3+1} = 2\). When \(x = 8\), \(u = \sqrt{8+1} = 3\). Thus, we have $$\int_2^3 \frac{\sin(u)}{u}(2u\, du).$$

Simplify the integrand

We get the integrand as $$\int_2^3\sin(u)\, du.$$

Evaluate the integral

Since the antiderivative of \(\sin(u)\) is \(-\cos(u)\), we have $$\left[-\cos(u)\right]_2^3 = -\cos(3) + \cos(2).$$ (ii) Evaluate \(\int_{0}^{\pi / 4} \cos 2 x \sqrt{4-\sin 2 x} d x\)

Substitution

Let \(u = \sin(2x)\). Then, \(du = 2\cos(2x)\, dx\). The integral becomes $$\int_{u(0)}^{u(\pi/4)} \frac{1}{2}\sqrt{4-u}\, du.$$

Calculate the new limits

When \(x = 0\), \(u = \sin(2\cdot 0) = 0\). When \(x = \pi/4\), \(u = \sin(\pi/2) = 1\). Thus, we have $$\int_0^1 \frac{1}{2}\sqrt{4-u}\, du.$$

Evaluate the integral

Since the antiderivative of \(\frac{1}{2}\sqrt{4-u}\) is \(\frac{2}{3}(4-u)^{3/2}\), we have $$\left[\frac{2}{3}(4-u)^{3/2}\right]_0^1 = \frac{2}{3}(4-1)^{3/2} - \frac{2}{3}(4-0)^{3/2} = \frac{2}{3}(3^{3/2} - 4^{3/2}).$$ (iii) Evaluate \(\int_{0}^{1-e^{-2}} \frac{\ln (1-t)}{1-t} \mathrm{dt}\)

Integrate by parts

Let \(u = \ln(1-t)\), which implies \(du = -\frac{dt}{1-t}\) and \(dv = \frac{dt}{1-t}\), which implies \(v= -\ln(1-t)\). We get $$\int_{0}^{1-e^{-2}} (-u)\, dv = uv\Big|_{0}^{1-e^{-2}} - \int_{0}^{1-e^{-2}} v \, du.$$

Evaluate the integrals

Since \(uv\Big|_{0}^{1-e^{-2}} = -\ln(1-t)^{2}\Big|_{0}^{1-e^{-2}}\), and $$-\int_{0}^{1-e^{-2}}(1-t)\ln(1-t)\, dt.$$ Finally, we need to compute the new integral.

Indefinite integral

Since it is difficult to find an elementary antiderivative for the remaining integral, we leave it as $$\int_{0}^{1-e^{-2}}(1-t)\ln(1-t)\, dt$$ and move on to the next question. (iv) Evaluate \(\int_{0}^{1} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)

Integrate by recognizing a derivative

Notice that the derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Thus, we get $$\int_{0}^{1} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x = \int_{0}^{1} (\sin^{-1}(x))'d x.$$

Evaluate the integral

Since the integral is the derivative of \(\sin^{-1}(x)\), we now have $$\left[\sin^{-1}(x)\right]_{0}^{1} = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}.$$